Chapter 1 Fracture Mechanics

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Chapter 1 Fracture Mechanics, 2nd ed. (2015) Solution Manual
1
CHAPTER 1
THEORY OF ELASTICITY
1.1 A thin sheet made of an aluminum alloy having E = 67 GPa, G = 256 GPa and also v = 1/3
was used for two dimensional surface strain measurements. The measurements provided
10.5 10 , 20 10 ,
5 5
 x   x xx yy   and 240 10 .
5
 x xy  Determine the corresponding stresses.
Solution:



















 





MPa
MPa
v
v
v
E
yy xx
xx yy
yy
xx
12.44
2.89
(1 )
2
 
 


MPa
v
E xy
xy 60.45
2(1 )





1.2 Determine (a) the principal stresses and strains and (b) the maximum shear stress for the case
described in Problem 1.1.
Solution:
(a) xy MPa MPa xx yy xx yy 4.28 60.60
2 2
2
2
1,2     






 


 
   

56.32 MPa  1 
64.88 MPa  2  
The principal strains are
5 3
2 2
1,2 4.75 10 1.20 10
2 2 2
 
   







 






 


 x x
xx yy xx yy xy     

3
1
1.15 10
  x
3
2
1.25 10
   x
(b) The maximum
xy MPa xx yy 60.60
2
2
2
max   






 
 
 

Chapter 1 Fracture Mechanics, 2nd ed. (2015) Solution Manual
2
1.3 Calculate the diameter of a 1-m long wire that supports a weight of 200 Newton. If the wire
stretches 2 mm, calculate the strain and the stress induced by the weight. Let E = 207 GPa.
Solution:
2 10 0.20%
10
2 3
3
  



x
mm
mm
l
l
o

mm
P
d 0.78 4
 

  E  414 MPa
P A ( / 4)d 200 N
2
     
1.4 Derive an expression for the local uniform strain across the neck of a round bar being loaded
in tension. Then, determine its magnitude if the original diameter is reduced 80%.
Solution:
Volume constancy:
AL  AoLo
   
2
ln L / L ln d / d
L
dL
o o
L
L
zz
o
   

d d  zz o
  2ln /
If 0.80 ,
o
d  d then
 2ln1/ 0.80  0.4463 zz

 zz  44.63%
1.5 The torsion of a bar containing a longitudinal sharp groove may be characterized by a
warping function of the type [after F.A McClintock, Proc. Inter. Conf. on Fracture of Metals,
Inst. of Mechanical Eng., London, (1956) 538] 
  
r
z w ydx xdy
0
  ( ) . The displacements are
 0  x
and  y  rz , where  and r are the angle of twist per unit length and the crack tip
radius, respectively. The polar coordinates have the origin at the tip of the groove, which has a
radius (R). Determine w, the shear strains rz  and .
 z
 In addition, predict the maximum of the
shear strain .
 z

Chapter 1 Fracture Mechanics, 2nd ed. (2015) Solution Manual
3
Solution:
Let x  Rcos and y  Rsin so that

 
 












y z
z
z r
1
dx  Rsind and dy  Rcosd  
r
R
R r
r
r z
2
1 2 
       





 
 
   

   
0
2
0
( ydx xdy) R d
r
z
If r  0 then   z   max  
 0 rz 
1.6 A cantilever beam having a cross-sectional area of 1.5 cm2
is fixed at the left-hand side and
loaded with a 100 Newton downward vertical force at the extreme end as shown in the figure
shown below. Determine the strain in the strain gage located at 8 cm from the fixed end of the
shown steel cantilever beam. The steel modulus of elasticity of is E = 207 GPa.
Solution:
FBD:
Thus,
2
1
3
1
6
12
2
bh
P X X
nh
P( X X )( h / )





 


 
Using Hooke’s law yields the elastic strain
E

 
 
4
2 2 2
2
2
1
4.64 10
/ )(3 10 )(0.50 10 )
6
(207 10
(6)(100 )(20 8 ) 10 6
9

 





 x
x N m x m x m
N m m x
Ebh
P X X

M
X – X1












 
  



1
1
0
0
M P X X
P X X M
M ( Moments)
12
3
2
bh I
h C
I
MC


 
X = 20 cm
b = 3 cm
P = 100 N
h = 0.5 cm
Strain gage – +
8 cm
Chapter 1 Fracture Mechanics, 2nd ed. (2015) Solution Manual
4
1.7 The stress-strain behavior of an annealed low-carbon steel (σys = 200 MPa and E = 207 GPa)
obeys the Hollomon equation with k = 530 MPa and n =0.25. (a) Plot the true and engineering
stress-strain curves. Calculate (b) the tensile strength ( ts ) and (c) the strain energy density up to
the instability point.
Solution:
(a) Plots
 
   
  
 
 
ln 1
1
t
t
 
    
n
n
t t
k
k
  
 
  

1 ln 1
  
 
n
k






1
ln 1
(b) The engineering yield strain and the true strain at the instability point on the true stress-strain
curve are, respectively
-3
3
4.3478×10
207 10
900
  
x MPa
MPa
E
ys
ys


 n  0.25 t
 @ Instability point
At the instability or ultimate tensile strength point (maximum),
  
 
1 exp( ) 1 exp(0.25) 0.28403
ln 1
1,200 0.25 848.53 0.25
max
      
 
  
t
t
n
k t MPa MPa
 
 
 
0 0.05 0.1 0.15 0.2 0.25 0.3
1000
750
500
250
0
0 0.05 0.10 0.15 0.20 0.25 0.30
Strain
Stress
True
Engineering
Yield Point
 ,   0.0043, 305.81 MPa ys  ys 
 0.25 ts 
0
  
1 0.25
1200 ln 1
0.25





 
0.25 1200 t t   
 n  0.25 t

@ Instability point
Chapter 1 Fracture Mechanics, 2nd ed. (2015) Solution Manual
5
 
 
374.77 /1 0.28403 291.87 MPa
/ 1
1
max
max
  
 
 
ts MPa
ts
ts

  
  
(c) The strain energy density (W) is just the area under the stress-strain curve. In general, the true
and engineering strain energy densities are
    
       
3
2
1 0.25 1 0.25 -3
2 -3
1 1
2
0
0
33550 335.50 /
0.28403 4.3478×10 33550
1 0.25
1,200
2
207,000 4.3478×10
2 1
MJ m
m
m
m
MN W .
W . MPa
n
E k W E d k d
W d d
t
t
n
ys
n
ts
n ys
t
Elastic Plastic
t
ts
ys
ys
ts
ys
ys
 
 

 


   


















 
 
 
 
 

   
   






and
  
 
  
  
  
W . . i
W
n
E k
d
k W E d
W d d
n
ts ys
ys
n
Elastic Plastic
ts
ys
ys
ts
ys
ys
0 99932 0 95718
ln 0.28403- 4.3478×10
0.25 1
1,200
2
207,000 4.3478×10
ln 1 1
1 2 1
ln 1
0.25 1 -3
2 -3
1
2
0
0
 







 
  

 


  




















 
 
 




  
   






Therefore, the engineering strain energy density can not be determined analytically using the
modified Hollomon equation. Instead, the solution can be achieved numerically. Thus,
  
 
3
0.28403
0.0043478
0.0043478 0.25
0
170 60 /
170 60
1
1200 ln 1
207000
W . MJ m
W d d . MPa




   



 
Wt  2W
Chapter 1 Fracture Mechanics, 2nd ed. (2015) Solution Manual
6
1.8 The figure below shows a schematic cross-sectional view of a pressure vessel (hollow
cylinder) subjected to internal and external pressures. Determine the stresses at a point Pr,  in
polar coordinates when (a)  0 Pi
and  0, Pi
(b)  0, Po
(c)  0 Pi
and (d) a  0 so that the
hollow cylinder becomes a solid cylinder. (e) Plot f (r)  rr  and f (r)  rr  . Let a  450 mm,
b  800 mm,  0 Pi
and P  40 MPa. The valid radius range must be 0.45 m  r  0.80 m . Use
the following Airy’s stress function
2
1 2 3   c  c ln r  c r
Along with the boundary conditions
P r b
P r a
rr o
rr i
   
   
0 @
0 @
r
r


 
 
Solution:
(a) Derivatives of 2
1 2 3   c  c ln r  c r
2 3
2
2
2
3
2
2
2
c
r
c
r
c r
r
c
r
  


 




0
0
2
2










0
2

 



r
From eq. (1.58),
0
1 1
2
2
1 1
2
2
2 3
2
2
2
2 3
2
2
2
2

 





  



 














 



r r r
c
r
c
r
c
r
c
r r r
r
rr
Chapter 1 Fracture Mechanics, 2nd ed. (2015) Solution Manual
7
Using the boundary conditions yields
2 3
2
2 3
2
2 3
2
2 3
2
2 2
2 2
c
r
c
c P
r
c
P
c
r
c
c P
r
c
P
rr o o
rr i i
      
      


(a)
Replace r for the proper radius and solve eq. (a) for 2
c and 3
c
2 3
2
2 3
2
2
2
c
r
c
P
c
r
c
P
o
i
  
  
from which
2
2
3
2 3
2
2
2
a
c
c P
c
a
c
P
i
i
  
  
and
2
2
3
2 3
2
2
2
b
c
c P
c
b
c
P
o
o
  
  
Then,
  o i
o i
o i
i o
P P
b a
a b
c
P
P P
b
c
a
c
b
c
P
a
c
P




  
  
2 2
2 2
2
2
2
2
2
2
2
2
2
&
   
2 2
2
2 2
2
3
2 2
2
2 2
2
2 2
2
2 2
2
3
2 2
2 2
2 2
2 2
2 2
2
3
2
2 1
1
2
b a
b P
b a
a P
c
b a
a
P
b a
a P
P
b a
a P
b a
a P
c
P P
b a
a b
P P P
b a
a b
b
P
b
c
c P
i o
o
i
o
i i
o o o i o i o
















 






   

     
  Po Pi
b a
a b
c 

 2 2
2 2
2 &
 
2 2
2 2
3
2 b a
a P b P
c
i o



Thus,
 
 
 
 
0
2 2 2
2 2
2 2
2 2
2 2 2
2 2
2 2
2 2


















r
i o o i
i o o i
rr
b a r
a b P P
b a
a P b P
b a r
a b P P
b a
a P b P
Chapter 1 Fracture Mechanics, 2nd ed. (2015) Solution Manual
8
(b) If  0, Po
then (c) If  0, Pi
then
0
1
1
2
2
2 2
2
2
2
2 2
2




























r
i
i
rr
r
b
b a
a P
r
b
b a
a P
0
1
1
2
2
2 2
2
2
2
2 2
2




























r
o
o
rr
r
a
b a
b P
r
a
b a
b P
(d) If a  0, then
 0
 
 





r
o
rr o
P
P
(e) The plot
0.4 0.5 0.6 0.7 0.8
100
75
50
25
0
-25
-50
(MPa)

Radius r (m)

 rr
Chapter 1 Fracture Mechanics, 2nd ed. (2015) Solution Manual
9
1.9 Consider an infinite plate with a central hole subjected to a remote uniform stress as shown in
Example 1.5. The boundary conditions for this loaded plate are (1) S x  x 
   and σy = τxy = 0
at r   and (2)   0    r r
at r  a. Use the following complex potentials [12]


z 
S
4
1 
2a
2
z
2
and 

z  
S
2
1 
a
2
z
2

3a
4
z
4
to determine σr
, σθ and τrθ (in polar coordinates).
Solution: The infinite plate with a central hole subjected to a remote uniform stress is
Use the following stress expressions
r    2

z  2

z  4 Re

z
  r  i2r  2z
z  

ze
i2
Recall that the Euler’s formula for z and z² are
z  rei  rcos  isin 
z  rei  rcos  isin
z
m  r
me
im  r
3
cosm  isinm
and their real and imaginary parts are
Re z  r cos & Re z
m  r
m cosm
Imz  rsin  & Imz
m  r
m sinm
Apply the Euler’s formula to the above complex potentials so that


z 
S
4
1 
2a
2
z
2

S
4
1 
2a
2
r
2
e
i2


z 
S
4
1 
2a
2
r
2
e
i2
4Re

z  S 1 
2a
2
r
2
cos2
(a) The first stress equation:
r    4 Re
z  S 1 
2a
2
r
2
cos2
Then,
(a)
Chapter 1 Fracture Mechanics, 2nd ed. (2015) Solution Manual
10


z 
S
4

a
2
2z
2


z  
2Sa2
2z
3

Sa2
z
3

Sa2
r
3
e
i3

Sa2
r
3
e
i3


z 
Sa2
r
3
cos3 (Real part)
Thus,
z

z  rei

Sa2
r
3
e
i3

Sa2
r
2
e
i4
(b) The second potential:


z  
S
2
1 
a
2
z
2

3a
4
z
4
 
S
2
1 
a
2
r
2
e
i2

3a
4
r
4
e
i4
Hence,
  r  i2r  2z

z  

ze
i2
  r  i2r  2
Sa2
r
2
e
i4
e
i2 
S
2
e
i2 
Sa2
e
i2
2r
2
e
i2

3Sa4
e
i2
2r
4
e
i4
  r  i2r  S
2a
2
r
2
e
i2  e
i2 
a
2
r
2

3a
4
r
4
e
i2
Real parts:
  r  S
2a
2
r
2
cos2  cos2 
a
2
r
2

3a
4
r
4
cos2
Add eqs. (a) and (c) and solve for σθ
 
S
2
1 
a
2
r
2

S
2
1 
3a
4
r
4
cos2
Use eq. (a) to solve for σr
r 
S
2
1 
a
2
r
2

S
2
1 
4a
2
r
2

3a
4
r
4
cos2
Using the imaginary part of eq. (b) yields shear stress τrθ as
i2r  S i
2a
2
r
2
sin2  isin2  i
3a
4
r
4
sin2
r  
S
2
1 
2a
2
r
2

3a
4
r
4
sin 2
At r = a
r  0
  S  2S cos2  S1  2 cos2
r  0
(b)
(c)
(d1)
(d2)
(d3)
(f1)
(f3)
(f2)
Chapter 1 Fracture Mechanics, 2nd ed. (2015) Solution Manual
11
1.10 Use the Cauchy-Riemann condition to show that (a) f(z) = 1/z is analytic and (b) its
derivative is f(z)= -1/z².
Solution:
(a) If z = x + iy, then
1
z 
1
xiy 
1
xiy
xiy
xiy 
xiy
x
2y
2
1
z 
x
x
2y
2

iy
x
2y
2
Let
u 
x
x
2y
2
v  
y
x
2y
2
Then
u
x

x
2y
2
x
x x
2y
2

x
2y
2
2

x
2y
22x
2
x
2y
2
2

y
2x
2
x
2y
2
2
u
y
 
2xy
x
2y
2
2
and
v
x

2xy
x
2  y
2

2
v
y

x
2  x
2
x
2  y
2

2
Thus,
u
y
 
v
x
Therefore, f(z) = 1/z is analytic because ∂u/∂y = ∂v/∂x.
(b) The derivative of f(z) = 1/z is
f

z 
d
dz 
1
z
 
u
x
 i
v
x
f

z 
y
2x
2
x
2y
2
2
 i
2xy
x
2y
2
2

yix
2
x
2y
2
2

iiyx 
2
i x
2y
2
2

xiy
2
x
2y
2
2
 
z
2
xiyxiy
2
 
z
2
zz
2
 
1
z
2
Chapter 1 Fracture Mechanics, 2nd ed. (2015) Solution Manual
12
1.12 Evaluate the Cauchy integral formula given below for the complex function when zo = π.
fzo  
1
2i

cosz
z
21
dz
Solution:
The integral has to be modified as
fzo  
1
2i

cosz
z
2  1
dz
fzo  
1
2i

cosz
z  1z  1
dz
fzo  
1

fz
z  1
dz

fz
z  1
dz  2ifzo 
where
fz 
cosz
z1
Then,

fz
z  1
dz  2ifzo 

fz
z  1
dz  2i
coszo
zo  1 zo

fz
z  1
dz  2i
cos
  1
 1. 52i
Chapter 2 Fracture Mechanics, 2nd ed. (2015) Solution Manual
1
CHAPTER 2
INTRODUCTION TO FRACTURE
MECHANICS
2.1 Show that the applied stress  0 as the crack tip radius  0. Explain
Solution:
From eq. (2.28),
a
and a 
 

 max  max 2
1
 2 
Therefore,  0 as  0. This means that the applied stress  0 since the material has
fractured or separated, at least into two pieces, and there is not a crack present so that  0. On
the other hand,  0 means that the existing crack is very sharp as in the case of a fatigue
crack.
2.2 For a Griffith crack case, crack propagation takes place if the strain energy satisfies the
inequality U(a) U(a  a) 2a , where a  crack extension and  is the surface energy.
Show that the crack driving force or the strain energy release rate at instability is .
a
U(a)
G



Solution:
Use the Taylor’s series to expand the left side term of the inequality. Thus,
…………..
3
3
( )
3
3!
2 1
2
( )
2
2!
( ) 1
1!
1
( ) ( )  


 


 


    a
a
U a
a
a
U a
a
a
U a
U a a U a (2.1)
Using the first two terms yields
a
a
U a
U a a U a 


   
( )
( ) ( )
(2.2)
a
a
U a
U a U a a 


    
( )
( ) ( ) (2.2)
Chapter 2 Fracture Mechanics, 2nd ed. (2015) Solution Manual
2
Then,
a a
a
U a
  


 2
( )
(2.3)
G
a
U a
 


 2
( )
(2.4)
Therefore,
a
U a G



( )
(2.5)
2.3 A 1mmx15mmx100mm steel strap has a 3-mm long central crack is loaded to failure.
Assume that the steel is brittle and has E  207,000 MPa, 1500 MPa,
ys   and
K 70 MPa m.
IC  Determine the critical stress ( )  c
and the critical strain energy release rate.
Solution:
a = 1.5 mm and a / w  0.10
The geometry correction factor:
( / ) tan   1.07 



  
w
a
a
w
f a w



The stress intensity factor
K a IC    f 
MPa
a
KIC
c
1020
2
 
 

The strain energy release rate
E
KIC
IC
2
G 
GIC  23.70 kPa m
2 GIC  23.70 kN / m
2.4 Suppose that a structure made of plates has one cracked plate. If the crack reaches a
critical size, will that plate fracture or the entire structure collapse? Explain.
Answer: If the structure does not have crack stoppers, the entire structure will collapse since the
cracked plate will be the source of structural instability.
Chapter 2 Fracture Mechanics, 2nd ed. (2015) Solution Manual
3
2.5 What is crack instability based on according to Griffith criterion?
Answer: It is based on the energy consumption creating new surfaces or in developing crack
extension. This energy is the surface energy 2.
2.6 Can the Griffith Theory be applied for a quenched steel containing 1.2%C, if a pennyshaped crack is detected?
Answer: Despite the quenched steel is brittle, the Griffith Theory applies only if a plastic zone
does not form at the crack tip.
2.7 Will the Irwin Theory (or modified Griffith Theory) be valid for a changing plastic zone size
as the crack advances?
Answer: No. It is valid if the plastic zone remains constant as it the case in many materials
under plane strain conditions.
2.8 What are the major roles of the surface energy and the stored elastic energies in a crack
growth situation?
Answer: The surface energy acts as a retarding force for crack growth and the stored elastic
energy acts to extend the crack. Thus, concurrent with crack growth is the recovery of elastic
strain energy by the relaxation of atomic bonds above and below the fracture plane.
2.9 What does happen to the elastic strain energy when crack growth occur?
Answer: It is released as the crack driving force for crack growth.
2.10 What does U/a = 0 mean?
Answer: Crack growth occurs at ,   f
 as shown in eq. (2.23), since the magnitude of crack
growth and the crack resistance force become equal. That is,
E
a
s
2
2
 
 
If
E
a
s
2
2
 
  , then fracture occurs and the fracture toughness without any plastic zone
deformation is
s
γ
c
G  2 at ,   f
 as shown in eq. (2.31) with p  0. Thus,
s
γ
c
G  2 is related
to an irreversible process of fracture.
Chapter 2 Fracture Mechanics, 2nd ed. (2015) Solution Manual
4
2.11 Derive eq. (2.28) starting with eq. (2.20).
Solution:
From eq. (2.20),
Multiplying this expression by  yields the stress intensity factor as per
eq. (2.28)
a
a
   

    max 2
1
K a I
  



 

a
a
Kt

 
max
max
2
1
2
Chapter 3 Fracture Mechanics, 2nd ed. (2015) Solution Manual
1
CHAPTER 3
LINEAR ELASTIC FRACTURE
MECHANICS
3.1 A steel strap 1-mm thick and 20-mm wide with a through-the-thickness central crack 4-mm
long is loaded to failure. (a) Determine the critical load if KIC = 80 MPa m for the strap
material. (b) Use the available correction factor,   f (a / w), for this crack configuration. Now,
do the following three comparisons to have a better understanding: Feddersen:
  c
 ( fraction) , Irwin:   c
 ( fraction) and Koiter-Benthem:   c
 ( fraction) .
Solution:
B = 1mm, w = 20mm, 2a = 4 mm, a/w = 0.10, KIC = 80 MPa m , K a f a w I c
    / ,
(a) Finite plate:
From eqs. (3.31) and (3.32) along with o o
a / w  0.10 rad.  (180 )(0.10) /  5.7296
Feddersen [12]: f    
a
w
a
w
 sec  .
 1025  c
 984 64 . MPa
Irwin [13]: f  
a
w
w
a
a
w









tan 1017 .  c
 992 38 . MPa
Koiter-Benthem [14]: f      
a
w a
w
a
w
a
w


 







1
1 2
1 1304 1021
2
. .  c
 988 49 . MPa
(b) Infinite plate approach [a/w  0 since f(a/w)  1].
 
MPa
m
MPa m
a
KIC
c 1009.25
2 10
80
3


 
  

Feddersen: 97.56% (infinite)  c   c
Irwin: 98.33% (infinite)  c   c
Koiter: 97.94% (infinite)  c   c
Therefore, the calculated  c values do not differ much since the plate dimensions are large
enough and the initial crack size is relatively small.
B=1 mm
P 2a w P
Chapter 3 Fracture Mechanics, 2nd ed. (2015) Solution Manual
2
3.2 A steel tension bar 8-mm thick and 50-mm wide with an initial single-edge crack of 10-mm
long is subjected to a uniaxial stress  = 140 MPa. (a) Determine the stress intensity factor KI
. If
KIC = 60 MPa m , is the crack stable? (b) Determine the critical crack size, and (c) determine the
critical load.
Solution:
 = P/A= 140 MPa, KIC = 60 MPa m , 




  
w
a
K a f
I  
B = 8 mm, w = 50 mm, a = 10 mm, x = a/w = 0.20, A = Bw
From Table 3.1, KI
   a and
         
2 3 4
  f x  1.12  0.231 x  10.55 x  21.71 x  30.38 x
  f x  0.20  1.37
(a) The applied stress intensity factor
KI = 34 MPa m
KIC = 60 MPa m
Therefore, the crack is stable because KI 5 mm (Not valid)
c) B  7 mm
3.4 (a) One guitar steel string has a miniature circumferential crack of 0.009 mm deep. This
implies that the radius ratio is almost unity, d / D  1. b) Another string has a localized miniature
surface crack (single-edge crack like) of 0.009 mm deep. Assume that both
strings are identical with an outer diameter of 0.28 mm. If a load of 49 N is
applied to the string when being tuned, will it break? Given properties:
K 15MPa m, 795MPa.
IC ys   
Solution:
(a) Circumferential crack
If mm
D d
a 0.009
2


 , then d  0.262 mm
MPa
x m
N
d
P
908.87
(0.262 10 )
4 (4)(49 )
2 3
  

 















 










    
2 3
0.36 0.73
8
3
2
1
2
1
( / )
D
d
D
d
D
d
d
D
d
D
 f d D
For d / D  0.9357,
  1.14


Chapter 3 Fracture Mechanics, 2nd ed. (2015) Solution Manual
4
K  MPa  m
K a
I
I
3
(1.14) 908.87 0.009 10
 


  
KI
 5.51 MPa m
(b) Surface edge crack
  1.12
KI
   a
K  MPa  m I
3
(1.12) 908.87 0.009 10
  
KI
 5.41 MPa m
Therefore, the strings will not break since KI  KIC in both cases. When d / D 1 and the crack
size is very small, either approach gives similar results.
3.5 A 7075-T6 aluminum alloy is loaded in tension. Initially the 10mmx100mmx500mm plate
has a 4-mm single-edge through–the-thickness crack. (a) Is this test valid? (b) Calculate the
maximum allowable tension stress this plate can support, (c) Is it necessary to correct KI due to
crack-tip plasticity? Why? or Why not? (d) Calculate the design stress and stress intensity factor
if the safety factor is 1.5. Data: ys = 586 MPa and KIC = 33 MPa m .
Solution:
(a) mm
K
B
ys
IC 2.5 7.93
2
 




 

and  0.0375
w
a
Plane strain condition holds because the actual thickness is greater than the required value.
(b) KI
   a ; a = 4 mm and w = 100 mm
 
2 3 4
1.12 0.231 10.55 21.71 30.38 





 





 










   
w
a
w
a
w
a
w
a
w
a  f
   f 0.04  1.13
w
a  f
Thus,
MPa
a
KIC
  261
 

Chapter 3 Fracture Mechanics, 2nd ed. (2015) Solution Manual
5
(c) Plastic zone correction:
mm
K
r
ys
IC 0.17
6
1
2
 




 
 
K a a a r eff     eff with eff   = 4.17 mm
    
K MPa m
K MPa x m
eff
eff
33.76
1.13 261 4.17 10 3




Therefore, there is no need for crack tip plasticity since KI,eff  KIC .
(d) MPa MPa
SF d 174
1.5
261
  


MPa m
MPa m
SF
K
K
IC
Id 22
1.5
33
  
3.6 A steel ship deck (30mm thick, 12m wide, and 20m long) is stressed in the manner shown
below. It is operated below its ductile-to-brittle transition temperature (with KIc = 28.3 MPa m ).
If a 65-mm long through–the-thickness central crack is present, calculate the tensile stress for
catastrophic failure. Compare this stress with the yield strength of 240 MPa for this steel.
Solution:
2a = 65mm   f a / w 1 since a/w 0
w = 12 m
L = 20m
P 2a P
T
KI
(MPa m )
or
CVN
Energy
(Joules)
Transition Range.
Chapter 3 Fracture Mechanics, 2nd ed. (2015) Solution Manual
6
K a IC     KIC  28.3 MPa m and  ys  240MPa
MPa
m
MPa m
a
KIC 88.57
10
2
65
28.3
3






 
 




Therefore,    ys and to assure structural integrity, the safety factor (SF) that can be used to
avoid fracture or crack propagation is in the order of
2.71
88.57
240
  
MPa
MPa SF ys


3.7 Show that  0
da
KI
for crack instability in a large plate under a remote tensile external stress.
Solution:
K a I
  
0
2
1
 
da a
dKI 
 since   0 and a  0
3.8 The plate below has an internal crack subjected to a pressure P on the crack surface. The
stress intensity factors at points A and B are
 


 dx
a x
a x
a
P
KA

 


 dx
a x
a x
a
P
KB

Use the principle of superposition to show that the total stress intensity factor is of the form
KI
 P a
Solution:
According to the principle of superposition, the total stress intensity factor is
 

  












  
2 2
2
a x
a dx dx P
a x
a x
a x
a x
a
P
KI KA KB
 
P
A
B
2a
Chapter 3 Fracture Mechanics, 2nd ed. (2015) Solution Manual
7
Furthermore,
2 2
2
a x
a
( a x )( a x )
( a x ) ( a x )
a x
a x
a x
a x
a x
a x
a x
a x


 
  












 
 a
x
ar
a x
dx
csin
2 2
and
a
a x
KI 2P arcsin


If x = a on the crack surface, then  
o
arcsin 1   / 2  90 and
a P a
P
a
x
ar
a
KI P 


 
   
2
2
2 sin
3.9 A pressure vessel is to be designed using the leak-before-break criterion based on the
circumferential wall stress and plane strain fracture toughness. The design stress is restricted by
the yield strength  ys and a safety factor (SF). Derive expressions for (a) the critical crack size
and (b) the maximum allowable pressure when the crack size is equals to the vessel thickness.
Solution:
Select some alloy having known KIC and  ys values. The alloy with the highest
ys
KIC

2
and
2








ys
KIC

values should be used for constructing the pressure vessel.
(a) The critical crack size
F
ys
d
IC c
S
K a


  


Thus,
2
2
1













 

ys
F IC
c
c
F
ys
IC
S K
a
a
S
K
  

 
Chapter 3 Fracture Mechanics, 2nd ed. (2015) Solution Manual
8
3.10 A stock of steel plates with GIC = 130 kJ/m², σys = 2,200 MPa, E = 207 GPa, v = 1/3 are
used to fabricate a cylindrical pressure vessel (di = 5 m and B = 25.4 mm). The vessel fractured
at a pressure of 20 MPa. Subsequent failure analysis revealed an internal semi-elliptical surface
crack of a = 2.5 mm and 2c = 10 mm. (a) Use a fracture mechanics approach to predict the
critical crack length this steel would tolerate. (b) Based on this catastrophic failure, another
vessel was constructed with do = 5.5 m and di = 5 m. Will this new vessel fracture at a pressure
of 20 MPa if there is an internal semi-elliptical surface crack having the same dimensions as in
part (a)?
Solution:
(a) The vessel is based on the thin-wall theory since the thickness is
mm
d x mm B
i 250
20
5 10
20
3
   (Requirement)
B  25.4 mm (Given) is less than the required thickness. So use the thin-wall theory.
Thus, the fracture hoop stress is
h 
Pidi
2B

20 MPa 5×103 mm
2 25.4 mm
h  1,968.50 MPa
Using eq. (2.34) yields the critical crack length
ac 
EGIC
 1  v
2
h
2
ac 
207×103 MPa 130×103 MPa.m
 1  1/32
 1,968.50 MPa 2
ac  2.49 mm  2.5 mm
(b) For the new vessel,
B 
do  di
2

5.6 m  5 m
2
 0.30 m
do
di

5.6
5
 1.12
Also,
h 
do/di
2
 1
do/di
2
 1
Pi  8.86Pi
h  8.8620 MPa  177.23 MPa
  h  Pi  9.86Pi  197.23 MPa
Chapter 3 Fracture Mechanics, 2nd ed. (2015) Solution Manual
9
Then,
a/2c  2.5/10  0.25
/ys  197.23/2,200  0.09
From eq. (3.44),
Q 

2
2
3
4

a
2c
2 2

7
33

ys
2
Q 

2
2
3
4
 0.25
2
2

7
33 0.09
2
Q  1.63
Thus, the applied stress intensity factor is
KI  
 a
Q
KI  197.23 MPa  2.5×103

1.63
KI  13.69 MPa m
From eq. (2.33) at fracture,
KIC 
EGIC
1  v
2
KIC 
207×103 MPa 130×103 MPa.m
1  1/3
2
KIC  174 MPa m
Therefore, the new vessel will not fracture because KI < KIC. 3.11 A cylindrical pressure vessel with B = 25.4 mm and di = 800 mm is subjected to an internal pressure Pi . The material has KIC  31 MPa m and σys = 600 MPa. (a) Use a safety factor to determine the actual pressure Pi . (b) Assume there exists a semi-elliptical surface crack with a = 5 mm and 2c = 25 mm and that a pressure surge occurs causing fracture of the vessel. Calculate the fracture internal pressure Pf . (c) Calculate the critical crack length. Solution: (a) Let's figure out which pressure vessel wall theory should be used. B  di 20  800 mm 20  40 mm Chapter 3 Fracture Mechanics, 2nd ed. (2015) Solution Manual 10 Therefore, the thin-wall theory should be used. The actual or applied pressure is h  Pidi 2B  ys SF Pi  2Bys SFdi  2 25.4 mm 600 MPa 2.5 800 mm Pi  15.24 MPa (b) Using the principle of superposition yields   h  Pi  Pidi 2B  Pi   di 2B  1 Pi  800 2×25.4  1 Pi   16.75Pi  16.75 15.24 MPa   255.24 MPa (Actua l) Also, a 2c  5 25  0.20  ys  255.24 600  0.43 a B  5 25.4  0.20 M  Mk  1 [See eq.(3.46) & Figure 3.6] From eq. (3.44), Q   2 2 3 4  a 2c 2 2  7 33  ys 2 Q   2 2 3 4  0.2 2 2  7 33 0.43 2 Q  1.50 From eq. (3.41), KIC  MMk  a Q  16.75Pf  a Q Now, the surge pressure is Pf  KIC 16.75 Q  a  31 MPa m 16.75 1.50  5×103 m Pf  18.09 MPa Chapter 3 Fracture Mechanics, 2nd ed. (2015) Solution Manual 11 (c) The critical crack length is KIC    a Q  16.75Pi  ac Q ac  Q  KIC 16.75Pi 2  1.5  31 MPa m 16.75  15.24 MPa 2 ac  7.04 mm 3.12 This is a problem that involves strength of materials and fracture mechanics. An AISI 4340 steel is used to design a cylindrical pressure having an inside diameter and an outside diameter of 6.35 cm and 12.07 cm, respectively. The hoop stress is not to exceed 80% of the yield strength of the material. (a) Is the structure a thin-wall vessel or a thick-wall pipe? (b) What is the internal pressure? (c) Assume that an internal semi-elliptical surface crack exist with a = 2 mm and 2c = 6 mm. Will the vessel fail? (d) Will you recommend steel for the pressure vessel? Why? or Why not? (e) What is the maximum crack length the AISI 4340 steel can tolerate? Explain. Solution: Internal Crack External Surface From Table 3.2, σys = 1,476 MPa and KIC = 81 MPa for AISI 4340 steel. (a) Using the diameters yields do di  12.07 6.35  1.09 B  do  di 2  12.07 cm  6.35 cm 2 B  2.86 cm Therefore, the pressure vessel is based on the thick-wall theory because do di  1.1 B  di 20  0.32 (b) From eq. (3.43e), h  do/di 2  1 do/di 2  1 Pi  0.8ys Pi  0.8ys do/di 2  1 do/di 2  1 di  6.35 m dO  12.07 m a 2c a  2 mm and 2c  6 mm Chapter 3 Fracture Mechanics, 2nd ed. (2015) Solution Manual 12 Thus, the internal pressure is Pi  0.8 1,476 MPa 1.09 2  1 1.09 2  1  Pi  101.51 MPa (c) Fracture mechanics:   h  Pi  0.8ys  Pi   0.81,476 MPa  101.51 MPa   1,282.30 MPa Thus, a/2c = 2/6 = 0.33, σ/σys = 0.87 and the shape factor becomes Q   2 2 3 4  a 2c 2 2  7 33  ys 2 Q   2 2 3 4  0.33 2 2  7 33 0.87 2 Q  1.66 Then, KI    a Q  1,282.30 MPa  2×103 m 1.66 KI  78.89 MPa m Therefore, the pressure vessel will not fail because KI > B = 15 mm, di = 2 m) is to be made out of a weldable steel alloy
having σys = 1,200 MPa and KIC = 85 MPa. If an embedded elliptical crack (2a = 5 mm and 2c =
16 mm) as shown below is perpendicular to the hoop stress, due to welding defects, the given
data correspond to the operating room temperature and the operating pressure is 8 MPa, then
calculate the applied stress intensity factor. Will the pressure vessel explode?
Solution:
do  di  2B  2 m  2  15  103 m  2.03 m
do
di

2.03
2
 1.02
B 
di
20 
2  103 mm
20  100 mm
The pressure vessel is of thin-wall type because do/di  1.1 and B  di/20. The hoop stress and
the stress intensity factor for the embedded elliptical crack are, respectively
where
Then,
The pressure vessel will not explode because KI a and x a, these stress functions become
x  Re Z 
S
2
z
z
2  a
2
 1 
S
2
x
x
2  a
2
 1
y  Re Z 
S
2
x
x
2  a
2
 1
xy  0
#
For large z = x + iy = x since y = 0, the elastic stresses far from the crack tip vanish when x >> a
along the x-axis. Therefore,
x  0
y  0
xy  0
4.8 Consider the elliptical crack shown below and assume that the crack is in the z-plane where
z    a and pz  p . Derive the stress equations using the given crack geometry and the
Westergaard complex method.
pz  p 
ζ
z
θ
a
x
iy
Solution:
Let us move the origin to crack tip. Then the coordinates of P can be expressed in complex
variable z    a . Use the Westergaard complex equation so that
Zz 

1  a/z
2

z
z
2  a
2
Z 
  a
  2a

a 1 

a
2a 1 

2a
Chapter 4 Fracture Mechanics, 2nd ed. (2015) Solution Manual
9
For |ζ| << a, Z  a 2a   a 2 Z      a 2 3/2 But ζ = re iθ and Zr,    a 2rei  3/2   a 2r 3/2 e i/2 Zr,   a 2r cos  2  isin  2 Z  r,   a 2rei  3/2    a 2r 3/2 e i3/2 Z  r,   a 2rei  3/2    a 2r 3/2 cos 3 2  isin 3 2 Then, y  Re Z  y ImZ  y   a 2r cos  2 1  sin  2 sin 3 2 and x   a 2r cos  2 1  sin  2 sin 3 2 xy   a 2r cos  2 sin  2 sin 3 2 4.9 Assume that an infinite plate contains a through-central crack along the x-axis. If the plate is subjected to a remote or infinite stress loading condition, S, y  y       0,   x xy   then use the Sanford [54] modified Irwin-Sih complex potential 2  z  Zz    z Here, γ′(z) may be defined by a Cauchy integral, Z(z) is the Westergaard complex function, ψ∗ (z) is a complex polynomial and z is the complex variable define as z = x + iy. Based on this information, determine a function for the stress intensity factor KI and expand ψ∗ (z) when n = 0 and 1, and z = a, half the crack length. Chapter 4 Fracture Mechanics, 2nd ed. (2015) Solution Manual 10 Solution: The Westergaard complex function and the complex polynomial are Zz  KI 2z  z   n0 m bnz n/2 The given complex function yields KI 2z  2  z  n0 m bnz n/2 For n = 0 and 1 and z = a, KI  2a 2  z  bo  b1z 1/2  Chapter 5 Fracture Mechanics, 2nd ed. (2015) Solution Manual 1 CHAPTER 5 CRACK TIP PLASTICITY 5.1 Use the inequality KIC  KI as a criterion for crack instability where KI is defined by Irwin’s plastic zone corrected expression for a finite size, to determine if a steel pressure vessel is susceptible to explode under   200 MPa hoop stress. The vessel contains an internal circular crack perpendicular to the hoop stress. If K 60 MPa m, IC  700 MPa,  ys  and the crack size is 20 mm, (a) determine the ASTM E399 thickness requirement and the minimum thickness to be used to prevent explosion, (b) Will crack propagation occur at 200 MPa? (c) Plot B (thickness) vs.   ys / for a = 10, 20, and 30 mm, and (d) Will the pressure vessel explode when the crack size is 30 mm? Why? or Why not? and e) When will the pressure vessel explode? Solution: (a) From eq. (5.12) along with   2  ,   2 1 0.5( / ) I ys K   a    Let KIC  KI so that 2 2                  ys I ys KIC K   2 2 2.5 2.5 Actual ys I ys KIC K ASTM                    Thus, 2 2.5 Actual ys I ASTM K B                                                             2 2 2 2 2 2 1 2.5 1 1 0.5( / ) 2.5 YS ys ys ys a a BASTM            Chapter 5 Fracture Mechanics, 2nd ed. (2015) Solution Manual 2 If the equality is used, then BASTM  Bmin is the minimum thickness required by the ASTM E399 standard testing method. Substituting values, such as /  0.285714,   ys a = 20 mm, and   2 / in the derived thickness expression, yields BASTM  Bmin  5.41 mm . (b) KI a1 0.5( / ys   32.56 MPa m 2        . Therefore, crack propagation will not occur because KI  KIC . In this case, the crack is stable. . (c) Plot                                               2 2 2 2 2 2 1 2.5 1 1 0.5( / ) 2.5 ys ys YS ys B a a B ASTM ASTM            (d) It will not explode since KI ac 1 0.5( / ys   40 MPa m 2        and KI  KIC . (e) Using     2 1 0.5 ( / KIC  ac    ys yields the critical or maximum crack size as ac  67.91mm . Therefore, the pressure vessel will explode when a a 67.91 mm.  c  0 20 40 60 80 100 120 140 160 0.2 0.4 0.6 0.8 1   ys / BASTM 20 10 a  30 mm Chapter 5 Fracture Mechanics, 2nd ed. (2015) Solution Manual 3 5.2 A project was carried out to measure the elastic-strain energy release rate as a function of normalized stress     ys / of large plates made out a hypothetical brittle solid. All specimens had a single-edge crack of 3-mm long. Plot the given data and do regression analysis on this data set. Determine (a) the maximum allowable   ys / ratio for GIC  30 kPa.m and (b) KIC in MPa m . Given data: v  0.3, MPa ys   900 and E  207 GPa .   ys / 0 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 G kPa m IC . 0 0.40 1.70 1.90 7.00 12.00 19.00 26.00 36.00 48.00 Solution: The required regression equation and the plot are given below and the 2 3 0.2322 7.3168 53.4440 14.9570                             ys ys ys GIC       (a) Combining eqs. (3.5) and (5.12) yields                 2 4 2 2 3 2 2 2 2 2 2 42.0980 21.0940 207,000 1.12 1 0.3 3 10 900 1 0.5 1 / 1 0.5 / x x MPa x m MPa x x G E v a G IC ys ys ys IC                 where  /  0.75 ys x   at G 30 kPa.m. IC  (b) The fracture stress and the plane strain fracture toughness       MPa m MPa MPa m v EG K MPa MPa IC IC 82.61 1 0.3 207,000 0.03 . 1 0.75 900 675 2 2         ——————————————————————————————————————— 0 10 20 30 40 50 0.2 0.4 0.6 0.8 1   ys / kPa m GIC . Chapter 5 Fracture Mechanics, 2nd ed. (2015) Solution Manual 4 5.3 Determine the critical crack length of problem 5.2. Solution:   ys / 0 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 G kPa m IC . 0 0.40 1.70 1.90 7.00 12.00 19.00 26.00 36.00 48.00 Combining eqs. (3.5) and (5.12) yields                 2 4 2 2 3 2 2 2 2 2 2 42.098 21.094 207,000 1.12 1 0.3 3 10 900 1 0.5 1 / 1 0.5 / x x MPa x m MPa x x G E v a G IC ys ys ys IC                 where  /  0.75 ys x   at G 30 kPa.m. IC  The fracture stress and the plane strain fracture toughness       MPa m MPa MPa m v EG K MPa MPa IC IC 82.61 1 0.3 207,000 0.03 . 1 0.75 900 675 2 2         Thus,        mm MPa K MPa m a K a ys IC c ys IC c 1 0.5 0.75 3 1.12 675 1 82.61 1 0.5 1 1 0.5 1 2 2 1 2 2 2                                                               ——————————————————————————————————————— 0 10 20 30 40 50 0.2 0.4 0.6 0.8 1   ys / kPa m GIC . Chapter 5 Fracture Mechanics, 2nd ed. (2015) Solution Manual 5 5.4 A large brittle plate containing a central crack 4-mm long is subjected to a tensile stress of 800 MPa. The material has KIC = 80 MPa m , ys = 1200 MPa and v = 0.30. Calculate (a) the applied KI , (b) the plastic zone size using the Von Mises yield criterion and prove that r = rmax when .   o Consider all calculations for plane stress and plane strain conditions, and (c) draw the entire plastic zone contour where the crack tip is the origin of the coordinates. Solution: (a) KI a (800 MPa) 2 10 m 63.41 MPa m 3         with   f (a / w) 1 The crack is stable since KI < KIC. (b) From eq. (5.52),               sin 1 cos 2 3 4 2 2 2 h K r ys I ;           For plane strain For plane stress 1 2 1 2 v h If  = 0, then r hKI ys  2 2 2 and        0.07 plane strain 0.44 plane stress mm mm r Using eq. (5.52) yields  h r K ys I                 sin 3cos 2 1 2 For maximum KI and critical plastic zone size r = rc, let   r  0 so that             86.94 For plane strain 70.53 For plane stress 3 cos sin 3cos 0 o o  o    h h o o o Thus,                   0.37 For plane strain 0.59 mm For plane stress sin 1 cos 2 3 4 2 2 2 max mm h K r o o ys I    Furthermore,        r K h I ys   4 3 2 sin cos Chapter 5 Fracture Mechanics, 2nd ed. (2015) Solution Manual 6               2 2 2 2 2 2 2 2 3 cos 3sin 4 3cos 3sin cos 4      h  K h r K ys I ys I          0.63 plane strain 0.44 plane stress o 2 2 mm r mm    Therefore, r = rmax at o since   2 2 0 r  in both cases. (c) The contours in polar coordinates are as per               sin 1 cos 2 3 4 2 2 2 h K r ys I ——————————————————————————————————————— 5.5 (a) Use the data given in Example 3.3 for a pressure vessel containing a semi-elliptical crack (Figure 3.6) to calculate Irwin’s and Dugdale’s (a) plastic zones, (b) KI using Kabayashi’s finite size correction factor() and plasticity correction factor. (c) Compare results and determine the percent error against each case. (d) Is it necessary to include a plastic correction factor? Explain. Solution: Data from Example 3.3: a = 3 mm, 2c = 10 mm, B = 6 mm,   420 MPa, 700 MPa,  ys  a/(2c) = 0.30, a/B = 0.50, K 60 MPa m, IC  /  0.60,   ys M = 1, Mk = 1.12, and Q = 1.70 . The correction factor and the stress intensity factor are, respectively  1.12M M / Q  0.127  k KI   a  5.18 MPa m -0.4 -0.2 0 0.2 0.4 -0.1 0.1 0.2 0.3 0.4 Crack Tip Plane Strain Plane stress r mm  Radians Chapter 5 Fracture Mechanics, 2nd ed. (2015) Solution Manual 7 Comparison: (a) Irwin’s Approach: Dugdale’s Approach:  1.12M M / Q  0.127  k  1.12M M / Q  0.127  k   mm a r / ys 0.54 2 2     Eq. (5.13)   mm a r / 2 ys 1.33 2 2     Eq. (5.24) (b) Using a finite correction factor K (a r) I     K (a r) I         2 1 0.5 / I ys K   a    Eq. (5.12)                   2 2 1 0.5 ys I K a     Eq.(5.25)     2 1 0.5 / KI   a    ys                   2 2 1 0.5 ys I K a     KI (Irwin's)  5.63 MPa m KI (Dugdale's)  6.22 MPa m Comparisons: KI (eq. 3.29)  5.18 MPa m < K (Irwin's) I < KI (Dugdale's)  6.22 MPa m Dugdale’s – Irwin’s Error  100%(6.22  5.63)/ 5.63  10.48% Irwin’s – eq. (3.29) Error  100%(5.63  5.18)/ 5.18  8.69% Dugdale’s – eq. (3.29) Error  100%(6.22  5.18)/ 5.18  20.08% Observe that eq. (3.29) does not include plasticity correction and it yields a smaller value than both Dugdales’s and Irwin’s expressions. The latter expression gives an error of 26.5%, which is slightly large. On the other hand, 8.3% error seems to be more acceptable for comparison purposes. This implies that a plastic zone correction may not be necessary. (c) Using Irwin’s approach yields a plastic zone of 0.54 mm for one side of the semi-ellipse and as a result, r << a and the plastic zone size could have been excluded. On the other hand, Dugdale’s approach gives a plastic zone of 1.33 mm, which is large enough to be excluded; therefore, a plasticity correction is needed to obtain more a accurate KI-result. ——————————————————————————————————————— Chapter 5 Fracture Mechanics, 2nd ed. (2015) Solution Manual 8 5.6 A 50- mm thick pressure vessel is to support a hoop stress of 300 MPa at room temperature under no action of corrosive agents. Assume that a semi-elliptical crack (Figure 3.6) is likely to develop on the inner surface with the major axis 2c = 40 mm and semi-minor axis a = 10 mm. A 300-M steel, which is normally used for airplane landing gear, is to be considered. Will crack propagation occur at 300 MPa hoop stress? Make sure you include the Irwin’s plastic zone correction in your calculations and explain if it is necessary to do so. Use the data below and select the suitable tempered steel. 300-M Steel (MPa)  ys (MPa m) KIC 650o Temper 300o Temper 1070 1740 152 65 Solution: a = 10 mm, 2c = 40 mm, B = 50 mm, a/2c = 0.25, a/B = 0.20 Mk = 1.02 (From Figure 3.6b)  1.12M k / Q since M < 0.5 [lower limit according to eq. (3.46)] From eq. (3.42),   2 2 0.212 / Q      ys                   / 2 0 2 / 2 0 2 2 2 2 1 sin 1 0.75sin    d  d c c a   0.13169 (From a Table of Elliptic Integral of the Second Kind) Thus,     2 1 0.50 / I ys K   a      2 / 2 ys a r    300-M Steel (650o C Temper) 300-M Steel (300o C Temper)  / ys  0.28  / ys  0.17 Q  1.72 Q  1.67   0.87   0.88 KI  MPa m  KIC 47.16 KI  MPa m  KIC 47.13 r = 0.39 mm r = 0.15 mm r << a = 10 mm r << a = 10 mm These results are similar and crack propagation will not occur since both steels have KI < KIC. Select 300-M Steel (650o C Temper) since it has a larger KIC value. The plastic zone correction was not necessary. Thus, I MPa m KIC K   a  46.26  . Both steels can be used for the sought application. In fact, the calculated KI values are similar for both steels. This is purely accidental. Anyway, the plastically corrected KI values are approximately 2% higher. Chapter 5 Fracture Mechanics, 2nd ed. (2015) Solution Manual 9 5.7 If localized plasticity is to be considered, explain the physical meaning of the following inequality t ys E a      2 . Solution: If the applied stress ( ) is unchanged, but causes crack growth, then t ys  is constant since  ys  constant and . t c    In this case, the crack length is the only changing variable that increases; therefore, the inequality holds. ——————————————————————————————————————— 5.8 Show that /( ), t ys r    where r is the plastic zone due to dislocation networks within the plastic zone area ahead of the crack tip. Solution: If E K ys I t   2  and , ys ys E    then, r K K ys ys ys I ys I t ys                            2 2 2 2 2 Thus, ys t r   2  . ——————————————————————————————————————— 5.9 Show that   2 / / t ys KI  ys    and give a reasonable interpretation of this equality. Solution: Apparently, t ys  /  is related to the crack size [ t ys  /   a] since   2 / KI  ys is associated with the plastic zone [r   /  ]. 2 KI  ys Therefore, r  / . t ys   Here  means “proportional.” ——————————————————————————————————————— 5.10 A large plate containing a through-the-thickness central crack of 2 ac  20 mm has E  207 GPa, 1,275 MPa,  ys  and  c  9.47 m at service temperature. Determine (a) the plane strain fracture toughness, (b) the design stress intensity factor for a safety factor (SF) of 2, (c) the critical fracture stress, and (d) the design service stress. Solution: (a) E K ys IC c   2  (c) K a MPa c IC c   /   282 KIC  50 MPa m (d) SF MPa d c    /  141 (b) MPa m SF K K IC ID   25 Chapter 5 Fracture Mechanics, 2nd ed. (2015) Solution Manual 10 5.11 Predict t  for glass using 4 2 2 a x E     Solution: Glass is a brittle material and there is no need for plasticity correction; therefore, 0 @ x a.  t   ——————————————————————————————————————— 5.12 Develop a  t expression for a von Mises material. Compare it with t  for a Tresca material under plane strain condition. Assume that crack growth occurs along the crack plane. Solution: von Mises  t  expression: Tresca  t  expression: From eq. (5.53), From eq. (5.59) with   0 , 2 2 2 ys I hK r   and   E 2 K r I t  2 2 2 ys KI r   and   E 2 K r I t    ys I t E h K von Mises    2 1/ 2 2    ys I t E K Tresca    2 2  Combining Mises and Tresca t t   yields     Mises  v Tresca Mises h Tresca t t t t     1 2 1/ 2    For Poisson’s ratio v  1, Mises Tresca t t    because Mises Tresca r  r ——————————————————————————————————————— 5.13 A material has E  70 GPa,  ys  500 MPa and v  1/ 3. It has to be used as a plate in a large structure. Non-destructive evaluation detects a central crack of 50 mm long. If the displacement at fracture is 0.007 mm and the plate width is three times the thickness, calculate (a) the crack tip opening displacement, (b) the plane strain fracture toughness, (c) the plane strain energy release rate, (d) the plate thickness and (e) What’s the safety factor being indirectly included in this elastic-plastic fracture mechanics approach? Assume plane strain conditions as per eq. (5.49) and a fracture load of 200 kN. Solution: Given data: a = 25 mm,  f  0.007 mm ,   3 and v  1/ 3. 500 MPa,  ys  P = 200 kN and E  70 GPa . (a) tc f mm x m 5 2 0.014 1.40 10      (b) From eq. (5.49) with   3  1.7321, which is defined below eq. (5.5), the plane strain fracture toughness ( KIC ), as per Irwin’s model, is Chapter 5 Fracture Mechanics, 2nd ed. (2015) Solution Manual 11 KIC tcE ys  31.40×10 m70×10 MPa500 MPa 25.82 MPa m 2 1 2 1 5 3           2 3 2 2 2 8.46 / 70 10 (1 ) 1 1/ 9 25.82 G kJ m x MPa MPa m E v KIC IC      (c)  c  KIC / a  92.13 MPa m and 2 3B P wB Pc c  c      mm x kN m P kN B c c 26.90 3 92.13 10 / 200 3 3 2     and w  3B  80.70 mm (Width) (d)  /  500 / 92.13  5.43 ys c SF   ——————————————————————————————————————— 5.14 Repeat problem 5.13 using eq. (5.41). Compare results. Solution: Data: a = 25 mm,  f  0.007 mm ,   3 and v  1/ 3. 500 MPa,  ys  P = 200 kN, E  70 GPa and c f mm x m 5 2 0.014 1.4 10        3  4v  5 / 3 For plane strain (a) c = ys IC ys IC E K E v K      9 8 4 ( 1)(1 ) 2 2         MPa m E x m x MPa MPa K c ys IC 41.61 8 9 1.4 10 70 10 500 8 9 5 3        (b)    2 2 3 2 2 2 0.022 / 22 / 70 10 (1 ) 1 1/ 9 41.61 G MJ m kJ m x MPa MPa m E v KIC IC       (c)      MPa x m MPa m a KIC c 148.48 25 10 41.61 3        2 3B P wB Pc c  c      mm x kN m P kN B c c 21.19 3 148.48 10 / 200 3 3 2     and w  3B  63.57 mm (Width) (d)   3.37 c ys SF   Problem 5.13 KIC  41.61 MPa m MPa c   148.48 B  21.19 mm SF 5.43 Problem 5.14 KIC  25.82 MPa m MPa c   92.13 B  26.90 mm SF  3.37 Chapter 5 Fracture Mechanics, 2nd ed. (2015) Solution Manual 12 5.15 A hypothetical large metallic plate containing a 10-mm central crack is 30-mm wide and 5- mm thick and mechanically loaded in tension. This plate has E  69 GPa, MPa, ys   500 v  1/ 3, and   0.3% (plane stress strain). Determine (a) t  , (b) G, and c)  as per Irwin, Dugdale, Burdekin, Rice, and Average equations, and compare the results. Solution: Data: w  30 mm, B  5 mm, v  1/ 3, a  5 mm , MPa E GPa ys   0.3%,   500 ,  69 Calculations:   B  5 mm0.003  0.015 mm and KI   a    5 2 I G  ys  1.5×10 m 500 MPa  0.0075 m.MPa  7.5  kJ / m   (a) In general, ys c ys I t E a E K     2 2   and a tE ys      Irwin–eq. (5.31):        MPa mm mm x MPa MPa a tE ys 161 4 5 0.015 69 10 500 4 3       Dugdale–eq. (5.32):        MPa mm mm x MPa MPa a tE ys 128 2 5 0.015 69 10 500 2 3         Burdekin–eq. (5.40):        MPa mm mm x MPa MPa a tE ys 182 5 0.015 69 10 500 3         Rice–eq. (5.41):       MPa mm mm x MPa MPa a tE ys 322 5 0.015 69 10 500 3       Rice equation yields  Rice  2 Irwin. Chapter 5 Fracture Mechanics, 2nd ed. (2015) Solution Manual 13 5.16 Determine (a) the critical crack tip opening displacement ( c  ), (b) the plastic zone size (r) and (c) the fracture stress ( )  c for a large aluminum alloy plate containing a central crack of 5-mm long. Use the data given below and assume that plane strain conditions exist. Data: K 25 MPa m, 500 MPa, IC   ys  and E  70 GPa. Solution: (a) From eq. (5.40),      mm x MPa MPa m E K ys IC c 0.018 70 10 500 25 3 2 2      (b) For plane strain condition, mm MPa K MPa m r ys IC 0.13 500 25 6 1 6 1 2 2                       From eq. (3.24),    MPa x m MPa m a KIC c 282 2.5 10 25 3        ——————————————————————————————————————— 5.17 Show that 2 2 8 t 1 t a E             for plane stress conditions, where  is the crack opening displacement (COD) and t  is the crack tip opening displacement (CTOD). Schematically, plot  t   f  for various  and fixed a values. Solution: From eq. (5.24), From eq. (5.23a), ar E t 2 4   (5.30)   2 2 4 2 a r x E      (5.29) 2 4 2         E t ar (a) 2 2 2 2 4 a ar r x E       (d) 2 2 4 1         E t a r (b) If a  x, then (e)   4 2 2 2 4 1         E t a r (c) 2 2 4 ar r E     (f) Chapter 5 Fracture Mechanics, 2nd ed. (2015) Solution Manual 14 Inserting eqs. (b) and (c) into (f) yields     2 2 4 2 2 2 4 4 2 2 2 8 1 2 4 1 2 4 1 4 4 t t t t t t a E E a E a E E                                                             The plot is ——————————————————————————————————————— 5.18 If the plane strain fracture toughness (KIC) and the yield strength (ys) of a 12-mm thick C(T) steel specimen are 71 MPa.m1/2 and 1,896 MPa, respectively, determine (a) the validity of the fracture mechanics tension test as per ASTM E399 for the plate containing a single-edge crack of 10-mm long at fracture, (b) the fracture stress if the plate is 20-mm wide, (c) the critical crack tip opening displacement, (d) the plastic zone size, and (e) interpret the results with regard to plane strain condition. Use a Poisson's ratio of 1/3 and assume that the elastic modulus of the steel 207 GPa. Solution: Increasing  a a fixed  o   t  Chapter 5 Fracture Mechanics, 2nd ed. (2015) Solution Manual 15 (a) Using eq. (3.5) yields the critical stress intensity factor The minimum size requirements can be computed using eq. (3.30). Thus, Therefore, the test is valid because a,B10,12  amin ,Bmin 3.51 mm and a/w  0.5 is within the 0.2  a/w 1 valid range. So proceed with the required calculations. (b) The fracture stress is determined from eq. (3.29) along with α=9.6591 since a/w=0.5 (Consult Table 3.1). Thus, (c) The crack-tip opening displacement is calculated using Rice’s equation, eq. (5.41), with   3  4v  5/ 3 and  11 v  32 / 9. Chapter 5 Fracture Mechanics, 2nd ed. (2015) Solution Manual 16 (d) The plastic zone can be calculated using eq. (5.53) along with the constant h = (1 – 2v) 2 = 1/9 (e) The above results suggests that the plate met the ASTM E399 size requirements because min min a,B  a ,B and 0.2  a/w 1. The C(T) specimen breaks in a brittle manner because both the plastic zone size and the crack-tip opening displacement are very small. ——————————————————————————————————————— 5.19 Assume that isotropic solid material having a single-edge crack is subjected to a remote tensile stress at room temperature. Let the properties of the material be 500 MPa,  ys  Poisson’s ratio v 1/ 3 and E  72 MPa. Let the applied stress intensity factor for mode I loading be K 20 MPa m. I  Excluding microstructural details and microscale defects, use the Tresca yielding criterion to derive (a) an expression for the critical plastic zone angle (c ) and its magnitude when minimum principal stresses are equal ( 2  3 ), (b) determine when  min  2 and  min   3 by knowing the value of c , (c) the plastic zone size at c and K 20 MPa m. I  The Tresca yielding criterion is based on the maximum shear stress reaching a critical or failure level. Hence, the definition of the maximum shear stress for this criterion is      ys  0.5 0.5 max  max  min  Here  max and  min are principal stresses and  ys is the monotonic tensile yield strength of a solid material. Let  max 1 and  min   2 or .  min  3 Solution: Needed equations for solving the problem. The principal stresses are Hence, Chapter 5 Fracture Mechanics, 2nd ed. (2015) Solution Manual 17 (a) The expression for the critical plastic zone angle (c ). For plane strain and c  conditions, If v = 1/3, then and the minimum principal stresses become (b) The minimum stresses. Let's calculate the values of  2 and  3 at   c and .   c For 0.67967 38.942 , o c     rad  say o   35 so that For 20 , o   Chapter 5 Fracture Mechanics, 2nd ed. (2015) Solution Manual 18 Therefore, the calculated values of the principal stress  2 and  3 at   c and fixed r indicate that .  min   3 For 0.67967 38.942 , o c     rad  say o   40 so that For 60 , o   Therefore, the calculated values of the principal stress  2 and  3 at  c and fixed r indicate that .  min  2 Chapter 5 Fracture Mechanics, 2nd ed. (2015) Solution Manual 19 (c) The plastic zone sizes for plane stress and plane strain at o c  38.942 are As expected, r for plane stress is greater than that for plane strain. Chapter 5 Fracture Mechanics, 2nd ed. (2015) Solution Manual 20 5.20 Consider a ductile steel plate containing a 50-mm through-thickness central crack subjected to a remote tensile stress of 40 MPa. If the yield strength of the steel is 300 MPa, then calculate KI as per LEFM, Irwin’s approximation and the Dugdale’s strip yield criterion. Compare results. Repeat all calculations for 290 MPa remote stress. Explain. Solution: Given data:   40 MPa and MPa ys   300 LEFM: KI a  MPa 10 m 11.21 MPa m 2 50 40 3              Irwin’s:         MPa m MPa m a K ys I 11.26 1 0.5 40 / 300 10 2 50 40 1 0.5 / 2 3 2                  Dugdale’s:     MPa m x x MPa m a K ys I 11.25 2 300 40 ln sec 8 10 2 50 40 2 ln sec 8 2 3 2                                               These results are similar. However, if   290 MPa, then KI LEFM: KI  81.27 MPa m Irwin’s: KI 115.18 MPa m Dugdale’s: KI 130.01 MPa m Therefore, KI values are dubious. See Figure 5.4 for comparing Irwin’s and Dugdale’s approximation schemes. Chapter 5 Fracture Mechanics, 2nd ed. (2015) Solution Manual 21 5.21 A single-edge SE(B) specimen with B = 20 mm, w = 40 mm is used to determine the critical CTOD tc  and . IC J See Example 5.4 for the specimen configuration and the load-displacement plot. Assume plane strain conditions and let 38 , max P  kN 800 MPa,  ys  E  207 GPa, v  0.3, KIC  60 MPa m B  20 mm; w  40 mm;  p 1.00 mm; S 150 mm; z 1.5 mm; a 12 mm Solution: From Example 5.4, The geometry correction factor  : x  a /w  12/ 40  0.3                      1.5212 2 1 2*0.3 1 0.3 3 0.3 1.99 0.3 1 0.3 2.15 3.93 0.3 2.7 0.3 2 1 2 1 3 1.99 1 2.15 3.93 2.7 3/ 2 1/ 2 2 3/ 2 1/ 2 2                  x x x x x x x The applied stress intensity factor as per LEFM is        MPa m x m x m N x m Bw P S KI 54.19 20 10 40 10 1.5212 38000 150 10 3 / 2 3 3 3 3 / 2 max        Then, CTOD becomes       x m mm MPa MPa MPa m m E K ys I te 8.866 10 0.009 2 800 207000 54.19 3 2 2                   mm mm mm mm mm mm mm mm w a a z w a p tp 0.47715 0.44 40 12 12 1.5 0.44 40 12 1 0.44 0.44              tc   te   tp  0.008866  0.47715  4.78 mm Thus, the J-integral is     3 2 JIC  m ys tc  2 800 MPa 4.78×10 m  7.65 MPa.m  7.65MN/m    Chapter 5 Fracture Mechanics, 2nd ed. (2015) Solution Manual 22 Chapter 6 Fracture Mechanics, 2nd Ed. (2015) Solution Manual 1 CHAPTER 6 THE ENERGY PRINCIPLE 6.1 Use Dugdale’s model for a fully developed plane stress yielding confined to a narrow plastic zone. Yielding is localized to a narrow size roughly equal to the sheet thickness (B). This is a fully elastic case, in which the plastic strain may be defines as ε δ /B,  t where t is the crack tip opening displacement. If the J-integral is defined by dJ  d , then show that ys t E a      2 2  Solution: If the strain energy density and the J-integral at yielding are defined by           0 0 W d dys t ys o J BW B  ysd        Thus,     t o ys o J B ysd d       Solving the J-integral yields t ys o ys t J  d       Then, ys ys I t I t ys I E a E K E K and E K J         2 2 2 2 2     Here,  is a correction factor. Chapter 6 Fracture Mechanics, 2nd Ed. (2015) Solution Manual 2 6.2 The crack tip opening displacement (t) for perfectly plastic solution to the Dugdale model was derived by Rice in 1966 [21] as                   0 0 2 log sec 1      G k a t where  o   ys and   Remote tensile stress (   o ), a  Crack size, and G = Shear modulus. Show that the path-independent J-integral is defined by          , 1 1 2 f a E k a J     Solution: Using Taylor’s series on         0 2 sec   and on log               0 2 sec   yields 0 2  y  and   2 … 1 24 5 2 sec 1 2 4 2 y y y y       Let x  secy so that                        … 1 1 3 1 1 1 log 2 3 x x x x x For x  0                                                                                2 2 2 2 2 2 2 4 2 2 4 2 1 2 1 1 2 1 2 sec 1 sec 1 2 1 1 log 2 o o y y y y y y x x x     If  1,  o  then 2 2 4           o  and 2 2 2 1 2 log sec                        o  o    Thus,   2 0 2 2 1           o t G k a      From eq. (6.66),     G k a G k a J o o t 8 1 2 2 1 2 2 2 0                     The elastic shear modulus of elasticity is  v E G   2 1 . Thus, J becomes         , 4 1 1 2 f a E k v a J     Chapter 6 Fracture Mechanics, 2nd Ed. (2015) Solution Manual 3 6.3 A bending test specimen made out of carbon steel showed a load-displacement behavior If the area under the P vs.  curve is 10 joules at the onset of crack growth, determine (a) the fracture toughness in terms of IC J as per ASTM E-813 Standard, (b) KIC and its validity as per ASTM E399 testing method, and (c) tc  [according to eq. (5.31) with   3 ] , f  and r . Explain the meaning of the results. Solution: (a)       kJ/m x m MPa m m joules Bb A J IC 32 32 10 . 25 10 25 10 2 2 10 2 3 3 3          (b) KIC EJ IC 207×10 MPa32×10 m.MPa 81.39 MPa m 3 3     Validity:   BASTM  KIC ys  mm  B BActual  BASTM 2.5 / 7.36 and min 2  Therefore, KIC  81.39 MPa m is valid. (c) From eq. (5.31),         mm x MPa MPa MPa m E K ys IC tc 0.016 3 207 10 1,500 4 4 81.39 3 2 2        Using eqs. (5.44) and (5.13) yields the fracture strain and the plastic zone size, respectively 6.4 10 0.064% 25 0.016 4 x or mm mm B c f       The plastic zone size: mm a MPa K MPa m r ys IC                     0.16 1,500 81.39 6 1 6 1 2 2    Therefore, the material behaved in a brittle manner implying that very little plastic deformation took place at the crack tip. These are very small values indicating that the material is brittle and undergoes very little plastic deformation. With regard to the critical crack tip displacement of 0.016 mm, it indicates that the crack tip displacement is very small; that is,  y   c / 2  0.008 mm . P Area  A = Area =10 Joules E = 207 GPa ys = 1,500 MPa B = 25 mm b = 25 mm Chapter 6 Fracture Mechanics, 2nd Ed. (2015) Solution Manual 4 6.4 If I ys t J    is used to determine the fracture toughness, will t  be a path-independent entity? Explain. Solution: Using Dugdale’s model, Figure 5.3 and eq. (6.33) yields           a r a I ds x J Wdy T     According to Figure 5.3, dy  0 from a to a  r and . T   y   ys Thus,     t ys o ys y y a r a y y I ys t dx d dx x J                         Therefore, both I J and t  are path-independent. Only the initial and final values of t  are needed to solve the integral. In fact, t  is just an opening as well as a path-independent entity. 6.5 Assume that crack growth occurs when I IC J  J , where J IC KIC / E 2 2  (1 ) . If a welldeveloped plastic flow occurs, will these expressions be valid? Explain. Solution: It is valid in the elastic regime where a  r. y x P P 2 y 2a r Plastic Zone Chapter 6 Fracture Mechanics, 2nd Ed. (2015) Solution Manual 5 6.6 A double cantilever beam (DCB) is slowly loaded in tension up 10 MN as shown schematically below. Assume that there is no rotation at the end of the beam and that the beam is made of isotropic steel having the following properties: KIC  47 MPa m and E  207 GPa. Will fracture occur? Solution: From eq. (6.42) along with   2 E'  E / 1  v and /12, 3 I  Bh             3 3 2 2 3 2 3 2 2 3 2 3 2 2 2 2 2 207,000 / 20 10 10 10 12 1 12 1 0.3 10 10 20 10 ' MN m x m x m x MN x m EB h v P a BE I P a GI          GI 5.28×10 MPa.m 3     2 3 2 1 0.3 207,000 5.28 10 . 1      MPa x MPa m v EG K I I KI  34.66 MPa m Therefore, fracture will not occur because KI  34.66 MPa m  KIC  47 MPa m B h h  P A B h P  o   M M  Pa M a P  10 kN   2 a = 20 mm B = 20 mm h = 10 mm Chapter 7 Fracture Mechanics, 2nd Ed. (2015) Solution Manual 1 CHAPTER 7 PLASTIC FRACTURE MECHANICS 7.1 Determine (a) the J-integral (J) and (b) the dJ/da for a hypothetical steel plate containing a central crack of 114 mm long loaded at 276 MPa and exposed to room temperature air. What does TJ < TR mean? Assume plane strain conditions and that the stainless steel obeys the Ramberg-Osgood relation with curve fitting parameters such as ' 1.69 and n  5.42. Explain the results based on the strain hardening effect. Data: E  206,850 MPa, 207 MPa,  ys    0.3 and 130 . 2 J IC  kJ/m Dimensions: w  4a  228 mm, L  2w Solution: (a) For a central crack, use Table 3.1 for  x  a/w  57/ 228  0.25 1 0.5 20.46 81.72 1 0.50.25 20.460.25 81.720.25 1.0635 2 4 6 2 4 6    x  x  x        sec 0.25  1.1892 I e K   a The effective crack length needed for calculating   I I e K  K a is given below. e o o I y a n K n n n r 2 2 1 1 1 1 1 1                                                      &   2 1 / 1  P Po     6   e o o e y a n n P P a a wr a 2 2 1 1 1 1 / 1                                      With   sec 0.25  1.1892      e e e e a mm x a a mm a 2 2 2 57 8.5597 10 207 1.1892 276 5.42 1 5.42 1 6 1 1 125.86/81.75 1 57                                 ae  62.336 mm  2w 2a  2L Chapter 7 Fracture Mechanics, 2nd Ed. (2015) Solution Manual 2 K a   MPa  x m I e 3 1.1892 276 62.336 10      KI  145.25 MPa m The elastic J-integral values that include the effects strain hardening is   x MPa m E K E K J I I e        2 2 2 2 2 2 9.2815 10 206,850 (1 ) (1 0.3 ) 145.25 '  2 Je  92.82 kPa.m  92.82 kJ/m Plastic Part (continued): Table 7.2 for a central crack ε σ /E 0.001 o  o  and   0.001 Po  4(w  a)σo / 3  81.75 MN/m & P 2wσ 2228×10 m276 MPa 125.86 MN/m 3     w  a  228  57  171mm The Plastic zone: r ry 10  17.855 mm 207 145.25 5.42 1 5.42 1 6 1 3 2                        From eq. (7.25) and Table 7.1 with n = 5.42, 6.2511 0.29381 1.1724 10 5.0031 2 2      I n n x n         3 5.42 1 3 1 10 207 276 ' 1.69 207 0.001 5.0031 17.855 10                   J  I r MPa x m n o p o o n      2 J p  87.17 kJ/m Thus, eq. (7.36) gives     2 2 J  Je ae  J p a,n  92.82 kJ/m  87.17 kJ/m 2 J  JI  180 kJ/m Therefore, crack grows occurs since 2 2 J  290.96 kJ/m  JIC  130 kJ/m (b) Crack instability   2 2 KI      E a E K E K J I I e (1  )  (1  )  ' 2 2 2 2      Chapter 7 Fracture Mechanics, 2nd Ed. (2015) Solution Manual 3 a n K n n n r r o o I y 2 2 1 1 1 1 1 1                                                       a n n J I r I n o o o o n n o p o o n 1 2 1 1 1 1 ' '                                                           a n n I E a J J J n o e p o o n 3 2 2 1 1 1 ' (1 )                                          0 1 1 1 ' (1 ) 3 2 2 2                            n o o o n n n I da E dJ                  5.42 3 2 2 2 207 276 5.42 1 5.42 1 6 1 1.69 207 0.001 5.0031 1.0635 206,850 (1 0.3 ) 1.1892 276                         MPa MPa x MPa da dJ  3 4.05 MPa 4.05 MJ/m da dJ   From eq. (6.70), the tearing modulus is     4.05  19.55 207 206,850 2 2   MPa  MPa MPa da E dJ T o J  TJ  19.55 Therefore, crack growth is stable if . TJ  TR Chapter 7 Fracture Mechanics, 2nd Ed. (2015) Solution Manual 4 7.2 Plot the given uniaxial stress-strain data for an ASTM A533B steel at 93ºC, and perform a regression analysis based on the Ramberg-Osgood equation. A single-edge-cracked plate made out of this steel is loaded in tension at 93ºC. Determine the total elastic-plastic J-integral when the applied load is   500 MPa . Will the crack grow in a stable manner? Why? or Why not? Assume plane strain conditions and necessary assumptions. Let a  100 mm, v = 0.30, E = 207 GPa, L = 3w, w = 400 mm, JIC = 1.20 MPa.m, 0.523 h1  and B = 150 mm. Strain (x10-3) 0 1.0 0 2.2 4 2.3 0 2.3 5 5.0 0 7.5 0 10 20 40 Stress (MPa) 0 381 414 415 416 450 469 483 519 557 Solution: Single-edge cracked specimen Data for plane strain condition: MPa ys o     414 (From the above Figure) a  100 mm B  150 mm E  207 GPa v  0.30 w  400 mm 2 1.20 6.8 in in kips J IC MPa m     (From Ref. [8]) L  3w  1,200 mm   500 MPa (Applied) b  w  a  300 mm (Ligament) Stress v s. Stra in 0 200 400 600 800 0 0.05 0.1 0.15 0.2 0.25 0.3 Stra in Stress (M Pa )  w a L  Chapter 7 Fracture Mechanics, 2nd Ed. (2015) Solution Manual 5 Elastic Part: Nonlinear regression using n          0 0 '      yields 1.12 '   and n  9.71. From Table 7.2, P wσ 400×10 m500 MPa 200 MN/m 3     100 0 82137 400 100 400 100 100 1 1 1/ 2 2 1/ 2 2 . w a a a w a a x                                        P 1.455xw a 1.4550.82137300×10 m414 MPa 148.43 MN/m 3 0   0     1  w  aa / w  400 100100 / 400  75 mm Also,   6 and 2 0 2 0 2 0 1 1 1 1 1 1 1 1                                                                                     n a a n n K n n n r I e e y   2 1 0 1 P/P w   The crack size can be calculated using       e e e e y a / . a a n n P/P a a a a wr                                                                 414 500 9.71 1 9.71 1 6 1 1 200 148 43 1 1 1 1 1 1 2 2 0 2 0    a . a a a . a e e e 10617 0 05814    a 100 mm ae 1.0617100 mm 106.17 mm For a /w  100/ 400  0.25, 1.12 0.23 /  10.55 /  21.71 /  30.38 /  1.5013 2 3 4    a w  a w  a w  a w  KI ae 1.5013500 MPa  106.17×10 m 469.10 MPa m 3             2 3 2 2 2 2 2 0 97 0 97 207 10 1 1 0 3 46910 . m.MPa . MJ/m x MPa . . MPa m E ν K E K J I ' I e        Chapter 7 Fracture Mechanics, 2nd Ed. (2015) Solution Manual 6 Plastic part: mm mm n a n r r e y 21 414 500 9.71 1 9.71 1 6 106.17 1 1 2 2 0                                                   From eq. (7.51) and Table 7.2, 0.523 h1  , 1.12 '   and mm x m 3 1 75 75 10          2 1 9.71 3 0 0 1 0 0 0 1 1 ' 0 89 148.43 200 1.12 414 0.002 75 10 0.523 0 002 J . MJ/m J MPa x m with ε σ /E . P P J h p p n p                           Thus, the total value of the J-integral is 2 J  Je  J p  0.97  0.89  1.86 MJ/m Therefore, crack will grow because J J 1.20 m.MPa.  IC  Stable? Unstable? Let’s find out how the crack will grow. Let I e K   a ;      E a E K E K J I I e e 2 2 2 2 ' 2 1   1      and 1 0 0 0 1 1 '           n p P P J     h so that the total value of the J-integral can be computed by    1 0 0 0 1 1 ' 2 2 1               n e e p P P h E a J J J        If a . a e  10617 and w aa / w 1   , then      w aa w P P . a h E J n 10617 / 1 1 0 0 0 1 ' 2 2                                  Thus, at constant load dJ/da is      MPa w a w w a P P h da E dJ n 1 26 1 1.0617 1 0 0 0 1 ' 2 2                                               26  31.40 414 207 10 2 3 2   MPa  MPa x MPa da E dJ T o J  Therefore, the crack will grow in a stable manner TJ   TR 31.40 Chapter 7 Fracture Mechanics, 2nd Ed. (2015) Solution Manual 7 7.3 Repeat Problem 7.2 using the Hollomon equation n   K for the plastic region. Curve fitting should be performed using this equation for obtaining K and n. Assume plane strain conditions and make the necessary assumptions. Compare the result from Problem 7.2. What can you conclude from these results? Assume a contour as shown in Figure 7.6 with y23 70mm. Solution: Single-edge cracked specimen Data for plane strain condition: MPa ys o     414 (From the above Figure) a  100 mm E  207 GPa w  400 mm v  0.30 B  150 mm 2 1.20 6.8 in in kips J IC MPa m     (From Ref. [8]) L  3w  1,200 mm   500 MPa (Applied) b  w  a  300 mm (Ligament) Elastic Part: For a /w  100/ 400  0.25, 1.12 0.23 /  10.55 /  21.71 /  30.38 /  1.5013 2 3 4    a w  a w  a w  a w  KI ae 1.5013500 MPa  106.17×10 m 469.10 MPa m 3             2 3 2 2 2 2 2 0 97 0 97 207 10 1 1 0 3 46910 . m.MPa . MJ/m x MPa . . MPa m E ν K E K J I ' I e        Stress v s. Stra in 0 200 400 600 800 0 0.05 0.1 0.15 0.2 0.25 0.3 Stra in Stress (M Pa ) Chapter 7 Fracture Mechanics, 2nd Ed. (2015) Solution Manual 8 Plastic part: Nonlinear regression using n   k yields k  776 MPa and n  0.103. The plastic strain energy density can be defined using the Hollomon equation as n   k and n k 1/          10.7087 1 1 0 0 776 500 1.103 776 1 1 776 0.103                                         MPa MPa MPa n K k n k W W d k d k MPa and n n n n p n p         3 Wp  6.35 MPa  6.35 MJ/m Assume a contour as shown in Figure 7.6 with y23 70mm so that p Wp J y23  2 as per eq. (7.43), which represents a near-field condition. Thus, J p 2y Wp 270×10 m6.35 MPa 0.89 m.MPa 3  23    2 J I  Je  J p  0.97  0.89  1.86 MPa  m  1.86 MJ/m Crack growth occurs since IC J  J . Note that if the value of 23 y changes so will p J and I J . Anyway, both problems 7.2 and 7.3 (P7.2 and P7.3) give the same result as per chosen contour segment 70 ; y23  mm otherwise J P7.2 J P7.3 p  p or J P7.2 J P7.3 I  I . Then, KI   a and   2 2 KI  a     E a E K J I e 2 2 ' 2  1    n n n p n K K n K W 1 1 1 1                     n n n p p n K K y n K J y W 1 23 1 23 1 2 1 2                         n n I e p n K K y E a J J J 1 23 2 2 1 2 1                             3 2 2 2 2 2 207 10 1 1 0.3 1.5013 500 x x da E dJ I          3 7.78 MPa 7.78 MJ/m da dJ I       7.78  9.40 414 207 10 2 3 2   MPa  MPa x MPa da E dJ T o J  Therefore, the crack will grow in a stable manner 9.40 . TJ   TR Chapter 7 Fracture Mechanics, 2nd Ed. (2015) Solution Manual 9 7.4 A steel plate having a single-edge crack is loaded as shown below. Calculate (a) the loadline displacement () and (b) the crack opening displacement ( ) that corresponds to a point on the resistance curve (not included). Assume plane strain conditions and use the following data: v  0.3, 0.523, h1  1.93, h2  3.42, h3  B  0.15 m, w  0.40 m, L  1.2 m and a  0.1 m. J . kJ/m , IC 2  120   500 MPa and E  206,850 MPa Solution: (a) From Table 7.2,   n p n p n p P a h P P P a h P P J w a h 1 0 0 0 31 0 3 1 0 0 0 1 ' ' so that '                                            Eliminating P/Po yields   n  n p p a h J w a h 1 / 0 3 0 0 1 ' '                      n  n p p n ah h w a J 1 / 3 1/ 0 0 1 '                  1 0 1 1 1/( 1) 3 0 '            n n n p p h J ah      Similarly,   1 0 1 1 1/( 1) 2 0 0 0 2 ' '                     n n n p n h J ah P P a h        These are the equations to be used in this problem. However, we need the value of p J at   500 MPa. Let’s get.  w a L  Chapter 7 Fracture Mechanics, 2nd Ed. (2015) Solution Manual 10 Data for plane strain condition: MPa ys o     414 (From the above Figure) a  100 mm B  150 mm E  207 GPa v  0.30 w  400 mm 2 1.20 6.8 in in kips J IC MPa m     (From Ref. [8]) L  3w  1,200 mm   500 MPa (Applied) b  w  a  300 mm (Ligament) Elastic Part: Nonlinear regression using n          0 0 '      yields 1.12 '   and n  9.71. From Table 7.2, P wσ 400×10 m500 MPa 200 MN/m 3     100 0 82137 400 100 400 100 100 1 1 1/ 2 2 1/ 2 2 . w a a a w a a x                                        P 1.455xw a 1.4550.82137300×10 m414 MPa 148.43 MN/m 3 0   0     1  w  aa / w  400 100100 / 400  75 mm Also,   6 and 2 0 2 0 2 0 1 1 1 1 1 1 1 1                                                                                   n a a n n K n n n r r I e e y   2 1 0 1 P/P w   The crack size can be calculated using       e e e e y a / . a a n n P/P a a a a wr                                                                 414 500 9.71 1 9.71 1 6 1 1 200 148 43 1 1 1 1 1 1 2 2 0 2 0    a . a a a . a e e e 10617 0 05814    a  100 mm ae 1.0617100 mm 106.17 mm Chapter 7 Fracture Mechanics, 2nd Ed. (2015) Solution Manual 11 For a /w  100/ 400  0.25, 1.12 0.23 /  10.55 /  21.71 /  30.38 /  1.5013 2 3 4    a w  a w  a w  a w  KI ae 1.5013500 MPa  106.17×10 m 469.10 MPa m 3             2 3 2 2 2 2 2 0 97 0 97 207 10 1 1 0 3 46910 . m.MPa . MJ/m x MPa . . MPa m E ν K E K J I ' I e        2 Je  0.97 MJ/m From Problem 7.2, mm mm n a n r r e y 21 414 500 9.71 1 9.71 1 6 106.17 1 1 2 2 0                                                   From eq. (7.51) and Table 7.2, P wσ 400×10 m500 MPa 200 MN/m 3     100 0 82137 400 100 400 100 100 1 1 1/ 2 2 1/ 2 2 . w a a a w a a x                                        P 1.455xw a 1.4550.82137300×10 m414 MPa 148.43 MN/m 3 0   0     1  w  aa / w  400 100100 / 400  75 mm 0.523 h1  , 1.12 '   and mm x m 3 1 75 75 10          1 9.71 3 0 0 1 0 0 0 1 1 ' 148.43 200 1.12 414 0.002 75 10 0.523 0 002                      J MPa x m with ε σ /E . P P J h p n p     2 J p  0.89 MJ/m 2 2 2 J  Je  J p  0.97 MJ/m  0.89 MJ/m  1.86 MJ/m Chapter 7 Fracture Mechanics, 2nd Ed. (2015) Solution Manual 12 Thus,            0.9066 0.0934 1 0 1 1 1/( 1) 3 0 414 0.3 0.523 0.89 . 0.1 3.42 1.12 0.002 '                   MPa m m MPa m h J ah p n n n p p        p  3.90 mm (b) The crack opening displacement is            0.9066 0.0934 1 0 1 1 1/( 1) 2 0 0 0 2 414 0.3 0.523 0.89 . 0.1 1.93 1.12 0.002 ' '                            MPa m m MPa m h J ah P P a h n n n p n           2.20 mm Therefore,  p  3.90 mm and   2.20 mm at the onset of crack growth since 2 J IC  1.2 MN/m and crack blunting required these values. 7.5 Determine the strain hardening exponent (n) for a steel with  ys = 400 MPa, E = 207 GPa. Assume that it obeys the Hollomon equation . n p   k Consider the maximum plastic stress in your calculations. Solution: max @ 1 max          p p and k MPa max    700 @ yielding, , n ys ys   k . max max n   k and 3 1 93 10   .  E ys ys       0.0895 ln 1.93 10 700 400 ln ln ln 3 max @ max max                      ys n ys n ys ys ys n k k          Chapter 7 Fracture Mechanics, 2nd Ed. (2015) Solution Manual 13 7.6 (a) Derive an expression for the J-integral J /J f(σ(σ ) p e  o ratio as per Rice model. (b) Plot the resultant expression for a remote stress to yield stress ratio range 0  /  1   o . Here,  o   ys . Explain the resultant trend. Solution: (a) The plastic case:                                      o o p o t o o t log sec G k a J log sec G k a            2 1 2 1 2 The elastic case:                                                     ( / ) 2 2 2log sec 8 1 8 1 2 2 2 o o o e p I e f J J G k a G k K J         From the plot, J p /Je   as 1 o σ/σ since secπ/ 2  and o σ  σ . 0 1 2 3 4 5 0.2 0.4 0.6 0.8 1 p e J /J o σ/σ They significantly differ from each other due to a large plasticity. At σ/σ 0.8 o  , JP dominates. Chapter 7 Fracture Mechanics, 2nd Ed. (2015) Solution Manual 14 7.7 Calculate the total J-integral (J) for a 2024 Al-alloy plate containing a single-edge crack under plane stress conditions. Plot b) the   f   and J f  . I  Use the following data to carry out all calculations: a = 1.40 mm, w = 19 mm, B = 0.8 mm, L = 10 cm, 64 MPa,  o  E  72,300 MPa, α′ = 0.35, n = 5 and F = 1.01 kN. Solution: Elastic part: a) The following calculations are self-contained. Thus, o   o E  64 72,300  8.852  104 A  wB  15.20×106 m2   F A  1. 01×103 MN 15.20×106 m2  66.45 MPa   ys For a single-edge crack, x  a/w  1.4/19  0.073684   1.12  0.230.073684  10.550.073684 2  21.710.073684 3  1.1516   1.12  0. 23x  10. 55x 2  21.71x 3  1.1516 Je  KI 2 E  a 2 2 E  1. 40×103 m1.1516 2 66.45 MPa 2 72, 300 MPa Je  3.5623  104 MPa.m  356.23 J/m2 Je  356. 23 J/m2 Plastic part: From Table 7.2 and eqs. (7.60), 1  w  aa/w  19  1.41.4/19  1.2968 mm   1   a wa  2  a wa  1  1.4 191.4 2  1.4 191.4  0. 92361 Po  1. 072w  a o  1. 072  0.92361  19  1. 4  103  64  1. 1153 MN/m P  w  19  103 m66.45 MPa  1.2626 MN/m h1  3a n 4 1n 1    o  n1 Po P n1 h1  31.4 5 41 1n 1.2968 66.45 64 51 1.1153 1.2626 51  3.3853 From eq. (7.54a), Jp   oo1h1 P Po n1  0.35  64  8.852  104  1.2968  103   3.3853 1.2626 1.1153 51 106  Jp  183.23 J/m2 JI  Je  Jp  356.23 J/m2  183.23 J/m2  539.46 J/m2 Chapter 7 Fracture Mechanics, 2nd Ed. (2015) Solution Manual 15 (b) These results indicate that J J p  0.34 and J 0.66J. e  Therefore, e p J  J and contributes 66% to J. (c) The required stress-strain curve can be determined using eq. (7.30). Thus,    ys   ys 1/n  322. 08 MPa 5  The J-integral plot along with the load point is based on the following equation: JI  2. 1247  10 2 2  1. 0638  10 9 6 The trend of the J f   I  resembles the trend shown in Figure 7.5. 0 10 20 30 40 50 60 70 80 0.0001 0.0002 0.0003 0.0004 Strain  Stress  (MPa) Load Point 80 60 40 20 0 Strain   -4  x10 0 1 2 3 4 0 20 40 60 80 200 150 100 50 0 JI (J/m 2 )  MPa Loading Point Chapter 7 Fracture Mechanics, 2nd Ed. (2015) Solution Manual 16 7.8 The plastic J-integral (Jp) for some configurations can be defined by J  ηW/(Bb), where W  absorbed energy, Bb = cross sectional area, b = (w – a) = ligament and   constant. This integral can then be separated into elastic and plastic components. For pure tension, Bb W E K J J J I p p e p      ' 2 (1) Consider a strain hardenable material and a specimen with unit thickness B. If the plastic load is defined by , n P  C p where C is the compliance and n is the strain hardening exponent, then the load-displacement and J-P profiles are schematically shown below. Derive an expression for .  p Solution: The area under the curve is the strain energy density define by p n p o p n p o p p P n n C n n W Pd C d p p        1 1 1          since n P  C p (2) Since Wp is a maximum energy then P reaches Po (the load limit). Thus, p Po p n n W  1  (3) From eq. (7.56), n o p o P P a h          3  '  and n o p o o P P P a h n n W           3 ' 1   (4) Substituting this expression into the J-integral equation above yields the plastic part as n o o o p p P P P a h n n Bb J           3 ' 1    (5)  W J P Po = Plimit Jlimit P Chapter 7 Fracture Mechanics, 2nd Ed. (2015) Solution Manual 17 Equate eq. (7.54) and (5) and solve for  p n o o o n o o o p P P h P P P a h n n b                   3 1 1 ' ' 1        3 1 2 1 h h a P b n n o o p     Chapter 8 Fracture Mechanics, 2nd Ed. (2015) Solution Manual 1 CHAPTER 8 MIXED-MODE FRACTURE MECHANICS 8.1 A large plate (2024-0 Al-alloy) containing a central crack is subjected to a combined mode III loading. The internal stresses are y = 138 MPa and xy = 103 MPa. Use the Maximum Principal Stress Criterion (-criterion) to calculate the fracture toughness (KIC). Use the following data: crack length 2a = 76 mm, = 1/3, and E = 72,300 MPa. Solution: a = 38 mm,  = 138 MPa, = 103 MPa KI   y a  47.68 MPa m KII  xy a  35.60 MPa m /  1.3393 KI KII Using eq. (8.29a) yields   o x x x x ar 48.60 9 3 8 1 3 1 cos 2 2 0                          From eq. (8.33), From eq. (8.34), 2 sin 2 3 cos 2 cos3 0 2 0 0 KIC KI  KII  KIIC KIC 62.77 MPa m 4 3    KIC  72.50 MPa m Chapter 8 Fracture Mechanics, 2nd Ed. (2015) Solution Manual 2 8.2 Repeat problem 8.1 using the Strain Energy Density Factor (S) Criterion (S-criterion). Solution: KI   a  47.68 MPa m KI  xy a  35.60 MPa m /  0.75 KII KI Use eq. (8.53) to solve for the fracture angle sin 2cos 1 4cos  1cos 2 sin 6cos 1 0 2 0 0 0 0 2 2 0 0           I I II KII   k K  k  K K   k o   48.10 Using Equation (8.54) yields From eq. (8.57) and (8.51),    2 11 2 2 12 22 2  1 1 4 IC a KI a KI KII a KII k E K       KIIC  0.9045KIC  53.68 MPa m KIC  59.35MPa m 2 Sc  14.26 kPa m  14.26 kJ / m 8.3 Determine (a) the incline angle  and (b) the applied tension  in Example 8.1. Solution: (a)    2 KI  a sin (b)    2 KI  a sin a = 38 mm KII   a sin cos 38 10 sin 53.25º 47.68 sin 2 3 2 m MPa m a KI          . MPa m . MPa m tan K K II I 35 60 47 68      214.95 MPa   53.25º 8.4 Calculate the critical stress (fracture stress) for the problem described in Problem 8.1 according to the  – criterion. Will crack propagation take place? Solution: From problem 8.1 & 8.2: From Example 8.4:   53.25º a = 38 mm o = -48.60º   215MPa (Applied) E = 72,300 MPa  = 1/3 Chapter 8 Fracture Mechanics, 2nd Ed. (2015) Solution Manual 3 From eq. (8.35), b b b  MPa a KIC c 2 cos cos 492 sin 1/ 2 2 11 12 22 0              where 1 cos  0.57314 8 1 3 11 0 b     sin 1 cos  0.77635 8 3 2 b12  0  0   sin 1 cos  1.05161 8 9 0 0 2 22 b      Crack propagation will not take place since   c  . 8.5 Repeat problem 8.4 using the S-criterion with   215 MPa. Solution:    2 KI  a sin    2 4 K  a sin I   sin cos 2  2 K K  a I II KII   a sin cos      2 2 2 K  a sin cos II From eq. (8.54),  1 2 2 12 22 2 11 2 4 1 1 / IC c o a sin a cos a cos E ( )( ) sin a K                   where        6 11 1 3 84 10 8 1 1 16 1          cos k cos . x E v cos k cos G a             6 12 2 1 1 15 10 8 1 2 1 16 1             sin cos k . x E v sin cos k G a            6 22 1 1 1 3 1 5 90 10 8 1          k cos cos cos . x E v a    Thus, MPa c   204   215 MPa (Applied) Therefore, crack propagation will occur because .   c  Chapter 8 Fracture Mechanics, 2nd Ed. (2015) Solution Manual 4 8.6 Show that the stress intensity factor is KI KIC  8 /11 when KI KIII  2 and that the Poisson’s ration is v  1/ 3. Solution: From eq. (8.13) with   2 E' E / 1 v and  0, KII              2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 8 11 4 1 1/ 3 1 1 4 1 1 1 1 4 1 1 1 (1 ) 1 (1 ) 2 '(1 ) IC I I I I I III IC I IC I III I III IC I III I III K K K K v v K K v K K K K v v K K E v E K K K with K K E E K K                                                Thus, KI KIC  11 8 Chapter 8 Fracture Mechanics, 2nd Ed. (2015) Solution Manual 5 8.7 Use the experimental mixed mode fracture data or Solid A and Solid B given below to compare the mixed mode fracture criteria discussed in this chapter. Determine which criterion is the most suited to predict the mixed mode fracture behavior of these solids. Solid A Solid B KII KI KII KI (MPa m  (MPa m  (MPa m  (MPa m  0.49 0 0.42 0 0.43 0.52 0.41 0.10 0.33 1.02 0.43 0.21 0.31 1.10 0.38 0.40 0.28 1.20 0.40 0.50 0.26 1.25 0.37 0.60 0.16 1.40 0.39 0.65 0 1.50 0.37 0.80 0.31 1.20 0.27 1.26 0.20 1.55 0 1.65 Solution: Plot the given data and use all possible mixed mode criteria in order to determine the most suited criterion. 0 0.125 0.25 0.375 0.5 2 1.5 1 0.5 0 Solid B Solid A MPa m KI K MPa m II Chapter 8 Fracture Mechanics, 2nd Ed. (2015) Solution Manual 6 From the given data, Solid A: KIC  1.50 MPa m & KIIC  0.49 MPa m Solid B: KIC  1.65 MPa m & KIIC  0.42 MPa m For Solid A, For Solid B, Therefore, the criterion 1 2 2                   IIC II IC I K K K K fits both data sets better than 1. 2           IIC II IC I K K K K 0 0.125 0.25 0.375 0.5 2 1.5 1 0.5 0 KII KI K MPa m II MPa m KI 1 2 2                   IIC II IC I K K K K 1 2           IIC II IC I K K K K 0 0.125 0.25 0.375 0.5 2 1.5 1 0.5 0 KII KI 1 2 2                   IIC II IC I K K K K 1 2           IIC II IC I K K K K K MPa m II MPa m KI Chapter 8 Fracture Mechanics, 2nd Ed. (2015) Solution Manual 7 8.8 A large and wide plate has a small through-the-thickness central crack as shown in the figure below. Use the free-body diagram (FBD) to derive expressions for (a) the normal (σN) and shear (τ) stresses and (b) the intensity factors KI and KII as functions of β. Solution: (a) Using the FBD yields the normal and shear stresses             2 2 cos sin cos 0 cos cos 0         N o o N y N o A A F A A                cos cos sin .cos cos 0 cos cos 0         o o x o A A F A A (b) The stress intensity factor equations are      2 K a sin K a I I N          K a sin cos K a II II   B  2a    Ao P    N A Area o  cos Ao A  FBD    Chapter 8 Fracture Mechanics, 2nd Ed. (2015) Solution Manual 8 8.9 Two identical cracked plates as shown in problem 8.8 are to be tested to determine KIC and KIIC . The observed critical tension loads and the incline angles were 120 MPa @   0 and 130 MPa @   / 4, respectively. Use the equations given below to determine the fracture toughness for mode I and II. 1 2           IIC II IC I K K K K and 1 2 2                   IIC II IC I K K K K Solution: For case   120 MPa @   0 and / 2 90 , o            K a     MPa m K a MPa x m MPa m II I sin cos 0 sin 120 20 10 1 30 2 3               1 2           IIC II IC I K K K K KIC  KI  30 MPa m For case   130 MPa @    / 4 and / 4 45 , o           K a      MPa  x m MPa m K a MPa x m MPa m II I 16.30 2 2 2 2 sin cos 130 20 10 16.30 2 2 sin 130 20 10 3 2 2 3                                          These results indicate that the care must be taken in selecting a particular fracture criterion. Nevertheless, the fracture criterion 1 2 2                   IIC II IC I K K K K gives conservative values for KIIC when compared to 1 2           IIC II IC I K K K K 0    K MPa m K MPa m K K K K K MPa m K K K K IIC IIC IC I IIC II IC IIC II IC I 24.12 16.30 1 16.30 / 30 1 1 30 1/ 2 1/ 2 2                                K MPa m K MPa m K K K K K MPa m K K K K IIC IIC IC I IIC II IC IIC II IC I 19.42 16.30 1 16.30 / 30 1 1; 30 1/ 2 2 1/ 2 2 2 2                                           Chapter 9 Fracture Mechanics, 2nd ed. (2015) Solution Manual 1 CHAPTER 9 FATIGUE CRACK GROWTH 9.1 Show that Paris equation can take the form   / 2 / 2 2 1 n t n ys v E A dN da              , where t  is the change of crack tip opening displacement, E is the modulus of elasticity, and A is a constant. Solution: From eq. (7.45) or (8.10) with KII = KIII = 0   E v K E K J I I 2 2 2 1 '    2 1 v EJ KI   2 1 v E J K     Substituting K into Paris expression, eq. (9.6) yields   n n v E J A K A dN da            2 1 From eq. (6.66), ys t J   and ys t J    . Thus,   / 2 / 2 2 1 n x n ys v E A dN da              9.2 (a) Show that   n C Ke dN da   , where   n( ) R A C    1 1 and    Ke  Kmax 1 R is Walker’s effective stress intensity factor range. (b) Plot   n C K dN da   and   n A K dN da   for a 4340 steel having 1,254 MPa,  ys  1,296 MPa,  ts  K MPa m, IC  130   0.42, n  3.24, R  0.7, and 5.11 10 . 11 A  x . Solution (a)  (1 ) Ke  K max  R max max max max max max min max min 1 K K K K K K K K K K K R           Chapter 9 Fracture Mechanics, 2nd ed. (2015) Solution Manual 2 (1 ) K  K max  R R K K    1 max Then,      1 (1 R) K Ke and   n n( ) n n e K ( R ) A ( R ) K A( K ) A dN da                 1 1 1 1 where n( ) ( R ) A C    1 1 For R  1 Thus,   n C K dN da   If R  0, then C  A. (b) If 1,254 MPa,  ys  1,296 MPa,  ts  K MPa m, IC  130   1/ 3, n  3, R  0.7 and cycles MN mm A x n n 2 2 2 11 5.11 10     , then n(1)  311/3  2 Thus,   x MN mm cycles x MN mm cycles R A C n 5.68 10 / 1 0.7 5.11 10 / (1 ) 10 3 4 2 11 3 4 (1 )           Comparison: Walker:   3 3 4 10 5.68 10 K cycles MN mm dN da             Paris:   3 3 4 11 5.11 10 K cycles MN mm dN da             Chapter 9 Fracture Mechanics, 2nd ed. (2015) Solution Manual 3 Plot vs. K dN da  for 20 MPa m  K  100 MPa m 9.3 Suppose that a single-edge crack in a plate grows from 2 mm to 10 mm at a constant loading frequency of 0.02 Hz. The applied stress ratio and the maximum stress are zero and 403 MPa, respectively. The material has a plane strain of 80 MPa.m1/2 and a crack growth behavior described by   12 4 3 68 10 KI . x dN da    Here, da / dN is in m/cycle and KI is in MPa m. Determine the time it takes for rupture to occur. Solution: KIC  80 MPa m ao  2 mm a f  10 mm R  0   1.12 f  20 Hz 12 3 68 10 A  . x n  4  0  min 403 MPa  max    403 MPa Using IC max c K   a yields ac  a f  10 mm 5e-09 1e-08 1.5e-08 2e-08 2.5e-08 0 20 40 60 80 100 mm cycles dN da / K mpA M  Walker Paris   a Chapter 9 Fracture Mechanics, 2nd ed. (2015) Solution Manual 4     4 A K A a dN da n   I    13,916 sec 0.020 /sec 278.31    cycles cycles f N t f        f f o a N a a da A dN 0 4 2    t  3.87 hours     N cycles m cycles x m x m N a a N f f o f f 278.31 1.51 . 2 10 10 10 1.51 1 1 1 3 1 3 1 1              9.4 If a large component is subjected to a cyclic loading under   300 MPa and R  0. The material behaves according to Paris Law   8 2 45 2 10 . KI x dN da    where da / dN is in m/cycles and KI is in MPa MPa m. Determine the plane strain fracture toughness for the component to endure 37,627 cycles so that a single-edge crack grows from 2 mm to a . c Solution:     2.45 2.45 8 2.45 2×10 K A1.12 a dN da   I     K   MPa  x m K a IC IC c 3 1.12 300 7 10 1.12             c f o a N a a da . dN , . 0 2.45 012577 4 732 30 KIC  50 MPa m 4,732.30 1.45 1.45 1.45     o c a a ac  a f  7 mm 9.5 Consider a part made of a polycrystalline metal that is stresses in the elastic stress range. If the metal contains inclusions, has an imperfectly smooth exterior surface, and natural dislocation, would the metal experience irreversible changes in a micro-scale? Explain. Answer: Many dislocations would move and contribute to a small irreversible process since the dislocation configurations would eventually change. This is one mechanism would lead to fatigue cracking due to dislocations. Chapter 9 Fracture Mechanics, 2nd ed. (2015) Solution Manual 5 9.6 Why most service fatigue fractures are normally not clear? Answer: Most service fatigue fractures experience varying stress amplitude, leading to irregularly spaced striations, if any. 9.7 What is the physical meaning of the slope n of the Stage II line according to the Paris model? Answer: According to Paris law,   n A KI dN da   , the higher the value of the slope n, the higher the crack growth rate da / dN. For instance, if K  0 and n  , then     n KI and da / dN   since A is a constant. 9.8 Suppose that d( 2a )/ dN  0.001 mm / cycle and n  4 in the Paris equation for 7075-T6 (FCC), 2024-T3 (FCC), Mo (BCC), and steel (BCC). Determine a) the constant A and its units, and b) which of these materials will have the higher crack growth, rate? Solution: From Figure 9.6,   2 4 A KI dN d( a )   and   4 3 10 KI mm / cycle A    Material KI MPa m ) A MN m / cycle 4 5   7075-T6 (FCC) 10 10 1 00 10 . x 2024-T3 (FCC) 15 11 1 98 10 . x Mo (BCC) 20 12 6 25 10 . x Steel (BCC) 30 12 1 23 10 . x Therefore, 7075-T6 (FCC) alloy has the highest fatigue crack growth rate because it has the highest constant A. Chapter 9 Fracture Mechanics, 2nd ed. (2015) Solution Manual 6 9.9 A Ti-6Al-4V large plate containing a 4-mm long central crack is subjected to a steady cyclic loading R = 0.10. The plate plane strain and the threshold fracture toughness are 70 MPa m and 14.7 MPa m. Determine (a) the minimum stress range, (b) the maximum applied stress range for a fatigue life of 3,000 cycles, and (c) the critical crack size for 3,000 cycles. Let the Paris equation be applicable so that n  4 and A MN m / cycles 12 4 12  10    . Solution: 2ao  4 mm N cycles 3  10 KIC  70 MPa m   1 ao  2 mm R  0.40 Kth  14.7 MPa m (a) Using eq. (9.2b) yields   . MPa x m . MPa m a K o th th 185 45 2 10 14 7 3          . MPa min th     185 45 (b) From eq. (9.9) with  0 No since o a exists,     0 3 2 1   2     C C C n n   ( n  2) C AN N n K  . m n o IC 2 0 50 1     C  ( / )( KIC / )  . x MPa  m 2 3 2 2 2   3 12 10 C ( a )( / ) ( K / ) . x MPa m n IC n / n / o    1 2 2 9 2 3 2 1   2 43 10 0 50 3 12 10 2 43 10 0 4 3 2 9 .   . x   . x  Thus, 4 2 5 2 1 3 2 2 1 1 2 2 4 3 12 10 3 81 10 2 1 2 C C C . x MPa . x MPa C C C           MPa   max 642 (Positive real root) (c) max min R    and    max  min R max   min and  min   max   R max   max   MPa . . MPa R max 713 1 0 1 642 31 1        Then, . mm K a max IC c 3 1 1 2             and 2ac  6.2 mm  2a  Chapter 9 Fracture Mechanics, 2nd ed. (2015) Solution Manual 7 9.10 Plot da / dN vs. K for 403 S.S. using the Paris, Forman and Broek/Schijve equations. Use the data given in Table 9.2 and a (20 mm)x(300 mm)x(900 mm) plate containing a single-edge crack of 2-mm long. Let 20 MPa m  K  80 MPa m. Solution: a x m 3 2 10  B  20 mm w  300 mm L  900 mm and a / w  0.007 and   f ( a / w)  1.12 Using data for AISI 403 stainless steel yields KIC  36 MPa m min max R   /  and    max  min Kth  1.37 MPa m R max   min and  min   max   MPa ys   1172 R max   max   R  0.1 n  3.5 / R  max   1 for R  1 13 5 98 10 A  . x Need to calculate K max for each given K. For instance, using the upper limit in the range 20 MPa m  K  80 MPa m, the procedure is as follows: a R K max max a        1 and K   a  80 MPa m Solve for the stress range,   901.12 MPa (Maximum)  / R MPa max    1  1001 (upper limit)  K max   max a  89 MPa m (upper limit)  . mm K a max IC c 48 30 1 2             It is suggested that you use a computer program to handle all the calculations needed in order to plot smooth curves. Paris :   da dN A K n   (9.6) Forman :     Plane strain  R K K A K = dN da IC n     1 (9.10) Broek/Schijve : da dN AK K n  max  2  (9.11) Chapter 9 Fracture Mechanics, 2nd ed. (2015) Solution Manual 8 9.11 Plot the data given below and use eq. (9.6) as the model to draw a curve fitting line on loglog scales. Determine the constants in such an equation. Solution: Since the Stage II data is for a linear behavior in a log-log scale, the constants in eq. (9.6) can be estimated as follows:   n A K dN da 1 1         and   n A K dN da 2 2         n K K dN da dN da                      2 1 2 1            20 30 2 14 10 8 84 10 8 8 1 2 1 2 ln / ln . x / . x ln K / K ln da / dN / da / dN n       n  3.50       3 50 8 1 1 20 2 14 10 n . MPa m . x m / cycles K da / dN A     A . x MPa m / cycles 13 3.50 8  5 98 10    Then, the equation for curve fitting is of the following nature   13 3 50 5 98 10 . . x K dN da    Perhaps, the curve fitting procedure gives slightly different results. No . MPa m K m / cycle da / dN 1 20 2.14×10-8 2 30 8.84×10-8 3 40 2.42×10-7 4 50 5.29×10-7 5 60 1.00×10-6 6 70 1.72×10-6 7 80 2.74×10-6 0 1e-06 2e-06 3e-06 4e-06 20 40 60 80 100 mm cycles dN da / K MPa m Chapter 9 Fracture Mechanics, 2nd ed. (2015) Solution Manual 9 9.12 A steel plate containing a single edge crack was subjected to a uniform stress range  at a stress ratio of zero. Fatigue fracture occurred when the total crack length was 0.03 m. Subsequent fatigue failure analysis revealed a striation spacing per unit cycle of 7.86×10-8 m. The hypothetical steel has a modulus of elasticity of 207 GPa. Predict (a) the maximum cyclic stress for a crack length of 0.01 m, (b) the striation spacing per unit cycle when the crack length is 0.02 m, (c) the Paris equation constants and (d) the plane strain fracture toughness and e) the fatigue crack growth rate assuming that the Paris equation is applicable nearly up to fracture. Solution: (a) x x m 8 7.86 10    max 8 8 23.69 6 7.86 10 207,000 6 7.86 10 / MPa m K x m MPa x K E x m cycles N a dN da a a                  Thus,   max 133.66 0.01 23.69           MPa m MPa m a K (b) For a  0.02 m , Kb  Kmax   a  133.69 MPa  0.02 m  33.50 MPa m Then, x m cycle cycle x m N x N a dN da x m MPa MPa m E K x b b 1.57 10 / 1 1.57 10 1.57 10 207,000 33.50 6 6 7 7 7 2 2                                   (c) The exponent in the Paris equation is     n b b n a a A K dN da A K dN da                     2 33.50 23.69 ln 1.57 10 7.86 10 ln ln / / ln 7 8                                x x K K da dN da dN n b a b a   a Chapter 9 Fracture Mechanics, 2nd ed. (2015) Solution Manual 10 and the constant is          10 2 10 2 2 8 1.40 10 1.40 10 / 23.69 / 7.86 10 / x K dN da x MPa cycle MPa m x m cycle K da dN A n a a           (d) For ac = 0.03 m,     K MPa m K a MPa m IC IC c 41 133.66 0.03 max      (e) Finally,        x K   x   x m cycle dN da x MPa cycle MPa m x m cycle K da dN A IC n a a 1.40 10 1.40 10 41 2.35 10 / 1.40 10 / 23.69 / 7.86 10 / 10 2 10 2 7 10 2 2 8              9.13 A 2-cm thick pressure vessel made of a high strength steel welded plates burst at an unknown pressure. Fractographic work using a scanning electron microscope (SEM) revealed a semielliptical fatigue surface crack (a = 0.1 cm and 2c = 0.2 cm) located perpendicular to the hoop stress and nearly in the center of one of the welded plates. The last fatigue band exhibited three striations having an average length of 0.34 mm at 10,000 magnification. The vessel internal diameter was 10 cm. Calculate a) the pressure that caused fracture and b) the time it took for fracture to occur due to pressure fluctuations. Assume a pressure frequency of 0.1 cycles per minute (cpm). Given data: ys = 600 MPa, E = 207 GPa, KIC  75 MPa m and da/dN = 4.50×10-7(K) 2 . Solution: 0.5 0.2 0.1 2   c a 0.05 2 0.1   B a x m cycle cycle m N a dN da 1.13 10 / 3 0.00034 4      Chapter 9 Fracture Mechanics, 2nd ed. (2015) Solution Manual 11 Using the given crack growth rate equation, K becomes K MPa m x MPa cycle x m cycle A da dN K 15.85 4.50 10 / / 1.13 10 / 1/ 2 2 4 1/ 2                      From eq. (9.23),      K MPa m m MPa x K E 15.58 6 10,000 0.00034 207,000 6      which is similar to the previous K value. The average becomes K 15.72 MPa m. Assume the following stress ratio /  0.64   ys so that   384 MPa, which is verified below. Thus, the shape factor becomes 0.64 2.38 33 7 0.2 0.1 4 3 33 2 7 4 2 3 2 2 2 2 2 2 2 2 2                                                            ys c a Q The hoop stress range along with the average value of K is   MPa m MPa m a Q K 386.32 1.12 0.001 / 2.38 15.72 /          Thus, 0.64 600 386.32     ys  Therefore, the assumed stress ratio is reasonably correct and fortunately, a single iteration is sufficient. The pressure is estimated as     MPa cm cm MPa d B P 154.40 10 2 2 2 384     (b) The critical crack length is x m cm x MPa K MPa m a IC c 9.56 10 0.956 1.12 386.32 1 1 75 3 2 2                         Chapter 9 Fracture Mechanics, 2nd ed. (2015) Solution Manual 12 Integrating eq. (9.15) along with  0 No yields                  cycles A x x a a N A a a a da A a A da A K da dN c o c o a a a a a a N N c o c o c o o 8.53 4.50 10 1.12 386.32 ln / ln(0.956/ 0.1) 1 ln / 2 7 2 2 2 2 2                               Then, 85.30 min 0.1 /min 8.53    cycle cycles cpm N t 9.14 Show that   2 max ac  a KIC / K Solution: From, KIC   max ac and K  a max  max a a K K a a a a K K IC c IC c c            2 max max max max     Then, 2 max          K K a a IC c Chapter 10 Fracture Mechanics, 2nd ed. (2015) Solution Manual 1 CHAPTER 10 FRACTURE TOUGHNESS CORRELATIONS 10.1 (a) Plot the given CVN–T data for a carbon steel. (b) Calculate KIC using the CVN values up to zero oC and Plot KIC-cal and KIC-exp. vs. Temperature. Is there a significant difference? If so, explain. T (oC) -125 -100 -60 -40 -12 0 10 25 40 45 60 80 CVN (J) 12 18 20 28 40 78 98 110 125 126 28 30 KIC-exp. (MPa.m1/2) 40 50 80 88 50 210 Solution: (a) Plots 0 20 40 60 80 100 120 140 -100 -50 0 50 100 T  C o C J  VN Chapter 10 Fracture Mechanics, 2nd ed. (2015) Solution Manual 2 (b) Do nonlinear curve fitting on KIC and CVN data set as per eqs. (10.66) and (10.67). Then, use eq. (10.68) for calculating KIC values. Thus, plot KIC-cal and KIC-exp. vs. Temperature. Below is a linear curve fitting result for plane strain fracture toughness and Charpy impact energy. 0 50 100 150 200 250 10 20 30 40 50 60 70 80 MPa m KIC U (J ) KIC  17.963  2.603U 20 40 60 80 100 120 140 160 180 200 220 240 -120 -100 -80 -60 -40 -20 0 MPa m KIC T  C o -2 2 -4 3 KIC  205.05 + 4.5721T + 5.0958×10 T + 2.0082×10 T Chapter 10 Fracture Mechanics, 2nd ed. (2015) Solution Manual 3 10.2 A mild steel plate has a through the thickness single-edge crack, a yield strength of 800 MPa and a static fracture strength is f  ys  3.2 . If the plate is loaded in tension and fractures at 600 MPa, calculate the plane strain fracture toughness of the steel plate and the critical crack length. Solution:  f  ys  3.2 and MPa ys   800   3.2 ys f    and 1/ 2 20    m From eq. (10.12),      MPa m m MPa K ys IC 88 20 1 3.2 1 800 1/ 2          If   600 MPa, then    mm MPa K MPa m a K a IC c IC c 5.46 1.12 600 1 88 1.12 1 1.12 2 2                      0 20 40 60 80 100 -120 -100 -80 -60 -40 -20 0 T  C o MPa m KIC Exp. Chapter 10 Fracture Mechanics, 2nd ed. (2015) Solution Manual 4 10.3 A standard Charpy specimen with B = 5 cm, w = 5 cm and S/w = 4 was tested at a room temperature. The measured impact energy was 30 joules. The tested material has a modulus of elasticity of 70,000 MPa. Calculate the KIC for this hypothetical specimen having x = a/w = 0.2. Solution: Data: x = a/w = 0.2 w = 5×10-2 m B = 5×10-2 m E = 70,000 MPa U = 30×10-6 MJ = 30×10-6 MN.m From eqs. (10.57) and (10.43), respectively,   0.45473 7 0.2 2 7 2       x        K MPa m x m x m x MN m MPa Bw UE K K IC IC I 42.98 0.45473 5 10 5 10 30 10 . 70,000 2 2 6         10.4 Suppose that a design code calls for a Charpy impact energy of 22 Joules for building a large pressure vessel containing an inert gas. If A533B and A723 ( MPa ys   1,100 ) steel plates are available for such a purpose, then (a) select the steel that will tolerate the largest critical crack length (depth of a surface semi-elliptical crack) when the hoop stress is 500 MPa and (b) determine the minimum plate thickness as per ASTM E399 standard for the selected steel. Solution: For 500 MPa, U 22 J and 1,100 MPa (A723),     ys    – . MPa m U ) K + – . IC 1 6550 196 A533B 20 139 0 54         eq. (10.71) KIC A723 0.644U ys 0.006 ys 91.24 MPa m 2      eq. (10.74) (a) The critical length mm MPa K MPa m a IC c 5.46 500 1 1 65.50 (A533B) 2 2                     mm MPa K MPa m a IC c 10.60 500 1 1 91.24 (A723) 2 2                     The ASTM A723 steel allows a larger critical crack length; therefore, this material is selected for such an application. (b) The plate thickness for the selected A723 steel is mm inches MPa K MPa m B ys IC 17.20 0.6772 1,100 91.24 2.5 2.5 2 2                      Chapter 10 Fracture Mechanics, 2nd ed. (2015) Solution Manual 5 10.5 Plot the fracture toughness data for a hypothetical polymer and determine the brittle-ductile transition temperature. T ( oC) -160 -120 -90 -80 -75 -70 -60 -50 -30 K (MPa m) IC 4.10 4.11 4.00 4.05 4.10 4.50 5.50 6.60 11.00 Solution: 10.6 For linear motion during Charpy impact tests, the energy lost by the striker (US ) and the kinetic energy ( 2 mv ) data for some polymers with different masses and spans to depth (S/w) ratios [47] are given below. Let m/M  1 and a) plot U f(mv²) S  and estimate the coefficient of restitution (e) for these data. What does 0  e  1 mean? Theoretically, determine b) the U f(mv²) S  ratio for the first bound which is transformed into specimen strain energy and c) the U f(mv²) S  ratio when there is no bouncing. mv2 mJ 0 50 100 150 200 US mJ 0 75 150 222 297 -200 -150 -100 -50 0 20 15 10 5 0 -200 -150 -100 -50 0 20 15 10 5 0 T  C O MPa m KIC -200 -150 -100 -50 0 20 15 10 5 0 -200 -150 -100 -50 0 20 15 10 5 0 -200 -150 -100 -50 0 20 15 10 5 0 -200 -150 -100 -50 0 20 15 10 5 0 T  C O MPa m KIC Brittle Ductile C o  80 Chapter 10 Fracture Mechanics, 2nd ed. (2015) Solution Manual 6 Solution: (a) The plot and the coefficient of restitution are Polynomial fit: 0 50 100 150 200 250 350 300 250 200 150 100 50 0 x y From eq. (10.39) along with m/M << 1,   0 5 1 1 5 1         e Slope e U e mv² S This result, e  0.5, means that a repeated impact of converging energy into strain energy occurs until the last contact is achieved. (b) The U f(mv²) S  ratio for the first bound which is transformed into specimen strain energy requires that the coefficient of restitution be e = 1. Thus, U /mv²  1 e  2 S c) The U f mv² S  ratio when there is no bouncing means that e  0 . Thus, U /mv²  1 e  1 S y  0. 6  1. 482x or US  0. 6  1. 482mv2  (mJ ) US mv ( ) 2 mJ Chapter 10 Fracture Mechanics, 2nd ed. (2015) Solution Manual 7 10.7 A large and thick ASTM A533B-1 steel plate containing a 4-mm long through-the-thickness central crack fractures when it is subjected to a tensile stress of 7 MPa. Plot U, KIC, and GIC at C T C o o  200   100 . Which of the plots is more suitable for determining the transition temperature? Explain. Data: E = 207,000 MPa and v = 1/3. Solution: From eq. (10.72) and (10.73), MPa m U K T U IC 1 in 196 20 139 in Joules 1 exp( 0.0297 ) 196 0.54             Denote that the transition temperature (To ) is easily determined using the U  f (T ) plot. This is shown with dash lines.           G    T  K T E v G IC IC IC 4.2941 20 95.50exp 0.016 20 95.50exp 0.016 in Joules 207,000 1 10 1 1/3 6 2 2 2        -200 -150 -100 -50 0 50 100 200 150 100 50 0 T U MPa m K U J IC ( ) T  C o T C o o  65 U KIC Chapter 10 Fracture Mechanics, 2nd ed. (2015) Solution Manual 8 It is clearly shown that the U f (T ) IC  plot is more suitable for determining the transition temperature T C o 65 0   . 10.8 A large plate made of 18Ni-8Co-3Mo Grade 200 alloy is part of structure exposed to relatively high temperature. Charpy impact tests were carried out and the average impact energy is 60 J. Use this information to calculate (a) the plane strain fracture toughness, (b) the minimum thickness ASTM requirement. The plate width is at least twice the thickness. Is this thickness practical? (c) Assume that a single-edge through the thickness crack develops. What will the critical crack length be? Will its value be reasonable? Solution: (a) If  60 J and MPa ys   1,310 (From Table 10.1), then eq. (10.76) yields KIC ys 2  4. 69 U ys  0.20 KIC  ys 4.69 U ys  0. 20 KIC  1, 310 MPa 4. 69 60 1,310  0. 20 KIC  159.42 MPa m -200 -150 -100 -50 0 50 100 4 3 2 1 0 T (C) GIC (J) (J ) GIC T  C o T C o 0  55 Chapter 10 Fracture Mechanics, 2nd ed. (2015) Solution Manual 9 (b) From eq. (3.30), the minimum plate thickness is B  2. 5 KIC ys 2  2.5 159. 42 1310 B  0. 30424 m  304.24 mm  12 inches This is a very impractical thick for plate, but it complies with the ASTM plane strain conditions. (c) Now if a single-edge crack develops during service, the critical crack length under a stress half the yield strength will be KIC  ys 2 ac ac  1  2KIC ys 2  103  2  159.42 1310 2 ac  19 mm This is a reasonable result for a plate so thick.

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